601_B. Lipshitz Sequence_21724
A function <image> is called Lipschitz continuous if there is a real constant K such that the inequality |f(x) - f(y)| ≤ K·|x - y| holds for all <image>. We'll deal with a more... discrete version of this term.

For an array <image>, we define it's Lipschitz constant <image> as follows:

  * if n < 2, <image>
  * if n ≥ 2, <image> over all 1 ≤ i < j ≤ n



In other words, <image> is the smallest non-negative integer such that |h[i] - h[j]| ≤ L·|i - j| holds for all 1 ≤ i, j ≤ n.

You are given an array <image> of size n and q queries of the form [l, r]. For each query, consider the subarray <image>; determine the sum of Lipschitz constants of all subarrays of <image>.

Input

The first line of the input contains two space-separated integers n and q (2 ≤ n ≤ 100 000 and 1 ≤ q ≤ 100) — the number of elements in array <image> and the number of queries respectively.

The second line contains n space-separated integers <image> (<image>).

The following q lines describe queries. The i-th of those lines contains two space-separated integers li and ri (1 ≤ li < ri ≤ n).

Output

Print the answers to all queries in the order in which they are given in the input. For the i-th query, print one line containing a single integer — the sum of Lipschitz constants of all subarrays of <image>.

Examples

Input

10 4
1 5 2 9 1 3 4 2 1 7
2 4
3 8
7 10
1 9


Output

17
82
23
210


Input

7 6
5 7 7 4 6 6 2
1 2
2 3
2 6
1 7
4 7
3 5


Output

2
0
22
59
16
8

Note

In the first query of the first sample, the Lipschitz constants of subarrays of <image> with length at least 2 are:

  * <image>
  * <image>
  * <image>



The answer to the query is their sum.
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
using Pii = pair<int, int>;
constexpr int MAX = 100010;
constexpr int INF = 0x3F3F3F3F;
unordered_map<int, int> mp;
int a[MAX], b[MAX], ll[MAX], rr[MAX], pre[MAX], stk[MAX], top;
i64 Solve(int l, int r) {
  int n = 0;
  for (int i = l; i < r; ++i) {
    b[++n] = abs(a[i] - a[i + 1]);
  }
  fill(pre, pre + n + 2, -1);
  b[0] = INF;
  b[n + 1] = INF;
  ll[0] = 0;
  top = 0;
  stk[++top] = 0;
  for (int i = 1; i <= n + 1; ++i) {
    for (; top && b[stk[top]] < b[i]; --top) {
      rr[stk[top]] = i;
    }
    if (b[stk[top]] == b[i]) {
      ll[i] = ll[stk[top]];
      pre[i] = stk[top];
    } else {
      ll[i] = stk[top];
    }
    stk[++top] = i;
  }
  i64 ret = 0;
  for (int i = 1; i <= n; ++i) {
    i64 lw = (i64)(i - max(ll[i], pre[i]));
    i64 rw = (i64)(rr[i] - i);
    ret += lw * rw * (i64)b[i];
  }
  return ret;
}
int main() {
  int n, q;
  while (~scanf("%d%d", &n, &q)) {
    for (int i = 0; i < n; ++i) {
      scanf("%d", &a[i]);
    }
    while (q--) {
      int l, r;
      scanf("%d%d", &l, &r);
      cout << Solve(l - 1, r - 1) << endl;
    }
  }
  return 0;
}
