280_D. k-Maximum Subsequence Sum_5219
Consider integer sequence a1, a2, ..., an. You should run queries of two types:

  * The query format is "0 i val". In reply to this query you should make the following assignment: ai = val. 
  * The query format is "1 l r k". In reply to this query you should print the maximum sum of at most k non-intersecting subsegments of sequence al, al + 1, ..., ar. Formally, you should choose at most k pairs of integers (x1, y1), (x2, y2), ..., (xt, yt) (l ≤ x1 ≤ y1 < x2 ≤ y2 < ... < xt ≤ yt ≤ r; t ≤ k) such that the sum ax1 + ax1 + 1 + ... + ay1 + ax2 + ax2 + 1 + ... + ay2 + ... + axt + axt + 1 + ... + ayt is as large as possible. Note that you should choose at most k subsegments. Particularly, you can choose 0 subsegments. In this case the described sum considered equal to zero. 

Input

The first line contains integer n (1 ≤ n ≤ 105), showing how many numbers the sequence has. The next line contains n integers a1, a2, ..., an (|ai| ≤ 500). 

The third line contains integer m (1 ≤ m ≤ 105) — the number of queries. The next m lines contain the queries in the format, given in the statement.

All changing queries fit into limits: 1 ≤ i ≤ n, |val| ≤ 500.

All queries to count the maximum sum of at most k non-intersecting subsegments fit into limits: 1 ≤ l ≤ r ≤ n, 1 ≤ k ≤ 20. It is guaranteed that the number of the queries to count the maximum sum of at most k non-intersecting subsegments doesn't exceed 10000.

Output

For each query to count the maximum sum of at most k non-intersecting subsegments print the reply — the maximum sum. Print the answers to the queries in the order, in which the queries follow in the input.

Examples

Input

9
9 -8 9 -1 -1 -1 9 -8 9
3
1 1 9 1
1 1 9 2
1 4 6 3


Output

17
25
0


Input

15
-4 8 -3 -10 10 4 -7 -7 0 -6 3 8 -10 7 2
15
1 3 9 2
1 6 12 1
0 6 5
0 10 -7
1 4 9 1
1 7 9 1
0 10 -3
1 4 10 2
1 3 13 2
1 4 11 2
0 15 -9
0 13 -9
0 11 -10
1 5 14 2
1 6 12 1


Output

14
11
15
0
15
26
18
23
8

Note

In the first query of the first example you can select a single pair (1, 9). So the described sum will be 17.

Look at the second query of the first example. How to choose two subsegments? (1, 3) and (7, 9)? Definitely not, the sum we could get from (1, 3) and (7, 9) is 20, against the optimal configuration (1, 7) and (9, 9) with 25.

The answer to the third query is 0, we prefer select nothing if all of the numbers in the given interval are negative.
#include <bits/stdc++.h>
using namespace std;
inline int Get_Int() {
  int Num = 0, Flag = 1;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') Flag = -Flag;
  } while (ch < '0' || ch > '9');
  do {
    Num = Num * 10 + ch - '0';
    ch = getchar();
  } while (ch >= '0' && ch <= '9');
  return Num * Flag;
}
namespace Segment_Tree {
class Data {
 public:
  int Left, Right;
  long long Sum;
  inline bool operator<(Data const &a) const { return Sum < a.Sum; }
  inline Data operator+(Data const &a) const {
    return (Data){Left, a.Right, Sum + a.Sum};
  }
};
class Node {
 public:
  Data LeftMax, RightMax, Max, LeftMin, RightMin, Min, Sum;
  inline Node operator+(Node const &a) const {
    Node x;
    x.LeftMax = max(LeftMax, Sum + a.LeftMax);
    x.RightMax = max(a.RightMax, RightMax + a.Sum);
    x.Max = max(x.LeftMax, x.RightMax);
    x.Max = max(x.Max, Max);
    x.Max = max(x.Max, a.Max);
    x.Max = max(x.Max, RightMax + a.LeftMax);
    x.LeftMin = min(LeftMin, Sum + a.LeftMin);
    x.RightMin = min(a.RightMin, RightMin + a.Sum);
    x.Min = min(x.LeftMin, x.RightMin);
    x.Min = min(x.Min, Min);
    x.Min = min(x.Min, a.Min);
    x.Min = min(x.Min, RightMin + a.LeftMin);
    x.Sum = Sum + a.Sum;
    return x;
  }
} A[200005 * 4];
bool Reverse[200005 * 4];
inline void Push_Down(int Now) {
  if (Reverse[Now]) {
    swap(A[Now].LeftMax, A[Now].LeftMin);
    swap(A[Now].RightMax, A[Now].RightMin);
    swap(A[Now].Min, A[Now].Max);
    A[Now].LeftMax.Sum *= -1;
    A[Now].RightMax.Sum *= -1;
    A[Now].Max.Sum *= -1;
    A[Now].LeftMin.Sum *= -1;
    A[Now].RightMin.Sum *= -1;
    A[Now].Min.Sum *= -1;
    A[Now].Sum.Sum *= -1;
    Reverse[Now << 1] ^= 1;
    Reverse[Now << 1 | 1] ^= 1;
    Reverse[Now] = false;
  }
}
inline void Update(int Now) {
  Push_Down(Now << 1);
  Push_Down(Now << 1 | 1);
  A[Now] = A[Now << 1] + A[Now << 1 | 1];
}
void Build(int Now, int *Value, int Left, int Right) {
  if (Left == Right) {
    A[Now].LeftMin = A[Now].RightMin = A[Now].Min = A[Now].LeftMax =
        A[Now].RightMax = A[Now].Max = A[Now].Sum =
            (Data){Left, Left, Value[Left]};
    return;
  }
  int Mid = Left + Right >> 1;
  Build(Now << 1, Value, (0 ? Mid + 1 : Left), (0 ? Right : Mid));
  Build(Now << 1 | 1, Value, (1 ? Mid + 1 : Left), (1 ? Right : Mid));
  Update(Now);
}
void Modify(int Now, int Position, int Value, int Left, int Right) {
  Push_Down(Now);
  if (Left == Right) {
    A[Now].LeftMin = A[Now].RightMin = A[Now].Min = A[Now].LeftMax =
        A[Now].RightMax = A[Now].Max = A[Now].Sum = (Data){Left, Left, Value};
    return;
  }
  int Mid = Left + Right >> 1, i = Position > Mid;
  Modify(Now << 1 | i, Position, Value, (i ? Mid + 1 : Left),
         (i ? Right : Mid));
  Update(Now);
}
void Flip(int Now, int left, int right, int Left, int Right) {
  Push_Down(Now);
  if (left == Left && right == Right) {
    Reverse[Now] ^= 1;
    return;
  }
  int Mid = Left + Right >> 1;
  if (left > Mid || right <= Mid) {
    int i = left > Mid;
    Flip(Now << 1 | i, left, right, (i ? Mid + 1 : Left), (i ? Right : Mid));
  } else {
    Flip(Now << 1, left, Mid, (0 ? Mid + 1 : Left), (0 ? Right : Mid));
    Flip(Now << 1 | 1, Mid + 1, right, (1 ? Mid + 1 : Left), (1 ? Right : Mid));
  }
  Update(Now);
}
Node Query(int Now, int left, int right, int Left, int Right) {
  Push_Down(Now);
  if (left == Left && right == Right) return A[Now];
  int Mid = Left + Right >> 1;
  if (left > Mid || right <= Mid) {
    int i = left > Mid;
    return Query(Now << 1 | i, left, right, (i ? Mid + 1 : Left),
                 (i ? Right : Mid));
  }
  return Query(Now << 1, left, Mid, (0 ? Mid + 1 : Left), (0 ? Right : Mid)) +
         Query(Now << 1 | 1, Mid + 1, right, (1 ? Mid + 1 : Left),
               (1 ? Right : Mid));
}
}  // namespace Segment_Tree
int N, Q, Top;
int Value[200005], temp1[200005], temp2[200005];
int main() {
  cin >> N;
  for (int i = 1; i <= N; ++i) Value[i] = Get_Int();
  Segment_Tree::Build(1, Value, 1, N);
  cin >> Q;
  while (Q--)
    if (Get_Int()) {
      int Left = Get_Int(), Right = Get_Int(), K = Get_Int();
      long long Ans = 0;
      while (K--) {
        Segment_Tree::Node Now = Segment_Tree::Query(1, Left, Right, 1, N);
        if (Now.Max.Sum <= 0) break;
        Ans += Now.Max.Sum;
        temp1[++Top] = Now.Max.Left;
        temp2[Top] = Now.Max.Right;
        Segment_Tree::Flip(1, Now.Max.Left, Now.Max.Right, 1, N);
      }
      while (Top) {
        Segment_Tree::Flip(1, temp1[Top], temp2[Top], 1, N);
        --Top;
      }
      printf("%lld\n", max(Ans, 0LL));
    } else {
      int x = Get_Int(), y = Get_Int();
      Segment_Tree::Modify(1, x, y, 1, N);
    }
  return 0;
}
