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 long long int MOD = 1000000007;
const long long BIG = 1446803456761533460;
const int Big = 336860180;
stringstream sss;
const long long int maxn = 100010;
long long int n;
long long int ar[maxn];
long long int num[maxn];
long long int cnt[maxn];
bool vis[maxn];
long long int c1 = 0, c2 = 0, odd = 0, ans = 0;
inline long long int R(long long int i) { return n - 1 - i; }
void MAIN() {
  cin >> n;
  for (long long int i = (0); i < (n); ++i) cin >> ar[i];
  for (long long int i = (0); i < (n); ++i) ++num[ar[i]];
  for (long long int i = (0); i < (n); ++i) odd += num[i + 1] % 2;
  if (odd > 1) {
    cout << "0\n";
    return;
  }
  for (long long int i = (0); i < (n / 2); ++i) {
    c2 += ar[i] != ar[R(i)];
  }
  long long int j = 0;
  for (long long int i = (0); i < (n); ++i) {
    while (j == i || j < n && (c1 > 0 || c2 > 0)) {
      if (j == R(j)) {
        c1 -= (++cnt[ar[j]] == 0);
      } else if (vis[R(j)]) {
        c1 -= (++cnt[ar[j]] == 0);
        c1 -= (++cnt[ar[j]] == 0);
      } else {
        c2 -= ar[j] != ar[R(j)];
        c1 += (cnt[ar[R(j)]]-- == 0);
        c1 -= (++cnt[ar[j]] == 0);
      }
      vis[j++] = true;
    }
    ans += n - j + (c1 <= 0 && c2 <= 0);
    if (i == R(i)) {
      c1 += (cnt[ar[i]]-- == 0);
    } else if (vis[R(i)]) {
      c1 += (cnt[ar[i]]-- == 0);
      c1 += (cnt[ar[i]]-- == 0);
    } else {
      c2 += ar[i] != ar[R(i)];
      c1 -= (++cnt[ar[R(i)]] == 0);
      c1 += (cnt[ar[i]]-- == 0);
    }
    vis[i] = false;
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(10);
  sss << R"(
6
3 6 5 3 3 5
5
5 5 2 5 2
    )";
  MAIN();
  return 0;
}
