504_C. Misha and Palindrome Degree_21711
Misha has an array of n integers indexed by integers from 1 to n. Let's define palindrome degree of array a as the number of such index pairs (l, r)(1 ≤ l ≤ r ≤ n), that the elements from the l-th to the r-th one inclusive can be rearranged in such a way that the whole array will be a palindrome. In other words, pair (l, r) should meet the condition that after some rearranging of numbers on positions from l to r, inclusive (it is allowed not to rearrange the numbers at all), for any 1 ≤ i ≤ n following condition holds: a[i] = a[n - i + 1]. 

Your task is to find the palindrome degree of Misha's array.

Input

The first line contains integer n (1 ≤ n ≤ 105).

The second line contains n positive integers a[i] (1 ≤ a[i] ≤ n), separated by spaces — the elements of Misha's array.

Output

In a single line print the answer to the problem.

Examples

Input

3
2 2 2


Output

6


Input

6
3 6 5 3 3 5


Output

0


Input

5
5 5 2 5 2


Output

4

Note

In the first sample test any possible pair (l, r) meets the condition.

In the third sample test following pairs (1, 3), (1, 4), (1, 5), (2, 5) meet the condition.
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int a[N], cnt[N];
int n;
bool isPali() {
  for (int i = 1; i <= n; i++) {
    if (a[i] != a[n + 1 - i]) return false;
  }
  return true;
}
bool canPali() {
  int odd = 0;
  for (int i = 1; i <= n; i++) {
    if (cnt[i] & 1) odd++;
  }
  return (n & 1) ? (odd <= 1) : (odd == 0);
}
bool ck(int x, int pos) {
  int mid = (1 + n) / 2;
  for (int i = mid; i > x; i--) {
    if (a[i] != a[n + 1 - i]) return false;
  }
  memset(cnt, 0, sizeof cnt);
  for (int i = pos; i <= x; i++) {
    cnt[a[i]]++;
  }
  for (int i = max(n + 1 - x, x + 1); i <= n + 1 - pos; i++) {
    cnt[a[i]]--;
  }
  for (int i = 1; i <= n; i++) {
    if (cnt[i] < 0) return false;
  }
  return true;
}
long long solve(int pos) {
  int l = pos, r = n + 1 - pos, ans;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (ck(mid, pos)) {
      ans = mid;
      r = mid - 1;
    } else {
      l = mid + 1;
    }
  }
  return (long long)pos * (n + 1 - ans);
}
long long solve() {
  int pos = 1;
  for (int i = 1; i <= n; i++) {
    if (a[i] != a[n + 1 - i]) {
      pos = i;
      break;
    }
  }
  long long ans = -(long long)(pos) * (pos);
  ans += solve(pos);
  reverse(a + 1, a + n + 1);
  ans += solve(pos);
  return ans;
}
int main() {
  while (scanf("%d", &n) == 1) {
    memset(cnt, 0, sizeof cnt);
    for (int i = 1; i <= n; i++) {
      scanf("%d", a + i);
      cnt[a[i]]++;
    }
    if (isPali()) {
      printf("%I64d\n", (long long)n * (n + 1) / 2);
    } else if (!canPali()) {
      puts("0");
    } else {
      printf("%I64d\n", solve());
    }
  }
  return 0;
}
