#include <bits/stdc++.h>
const int maxN = 100005;
int book[maxN] = {0}, sumOdd = {0}, num[maxN], count[maxN] = {0}, ans = 0;
int main() {
  int n;
  scanf("%d", &n);
  int l = 1, r = n;
  double mid = (1.0 + n) / 2;
  for (int i = 1; i <= n; i++) {
    scanf("%d", num + i);
    book[num[i]]++;
    if (book[num[i]] % 2)
      sumOdd++;
    else
      sumOdd--;
  }
  if (sumOdd > 1) {
    printf("0\n");
    return 0;
  }
  while (num[l] == num[r] && l <= r) {
    book[num[l]] -= 2;
    l++;
    r--;
  }
  if (l > r)
    printf("%I64d\n", (long long)(1 + n) * n / 2);
  else {
    for (int i = l; l <= r; i++) {
      if ((double)i < mid) {
        count[num[i]]++;
        if (count[num[i]] <= book[num[i]] / 2)
          ans++;
        else
          break;
      } else {
        if (num[i] == num[n + 1 - i] && (book[num[i]] % 2 || n % 2 == 0))
          ans++;
        else
          break;
      }
    }
    memset(count, 0, sizeof(count));
    for (int i = r; i >= l; i--) {
      if ((double)i > mid) {
        count[num[i]]++;
        if (count[num[i]] <= book[num[i]] / 2)
          ans++;
        else
          break;
      } else {
        if (num[i] == num[n + 1 - i] && (book[num[i]] % 2 || n % 2 == 0))
          ans++;
        else
          break;
      }
    }
    printf("%I64d\n", (long long)l * (l + ans));
  }
  return 0;
}

### Prompt

In CPP, your task is to solve the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
const int maxN = 100005;
int book[maxN] = {0}, sumOdd = {0}, num[maxN], count[maxN] = {0}, ans = 0;
int main() {
  int n;
  scanf("%d", &n);
  int l = 1, r = n;
  double mid = (1.0 + n) / 2;
  for (int i = 1; i <= n; i++) {
    scanf("%d", num + i);
    book[num[i]]++;
    if (book[num[i]] % 2)
      sumOdd++;
    else
      sumOdd--;
  }
  if (sumOdd > 1) {
    printf("0\n");
    return 0;
  }
  while (num[l] == num[r] && l <= r) {
    book[num[l]] -= 2;
    l++;
    r--;
  }
  if (l > r)
    printf("%I64d\n", (long long)(1 + n) * n / 2);
  else {
    for (int i = l; l <= r; i++) {
      if ((double)i < mid) {
        count[num[i]]++;
        if (count[num[i]] <= book[num[i]] / 2)
          ans++;
        else
          break;
      } else {
        if (num[i] == num[n + 1 - i] && (book[num[i]] % 2 || n % 2 == 0))
          ans++;
        else
          break;
      }
    }
    memset(count, 0, sizeof(count));
    for (int i = r; i >= l; i--) {
      if ((double)i > mid) {
        count[num[i]]++;
        if (count[num[i]] <= book[num[i]] / 2)
          ans++;
        else
          break;
      } else {
        if (num[i] == num[n + 1 - i] && (book[num[i]] % 2 || n % 2 == 0))
          ans++;
        else
          break;
      }
    }
    printf("%I64d\n", (long long)l * (l + ans));
  }
  return 0;
}

```