You are given a permutation p of n integers 1, 2, ..., n (a permutation is an array where each element from 1 to n occurs exactly once).

Let's call some subsegment p[l, r] of this permutation special if p_l + p_r = max _{i = l}^{r} p_i. Please calculate the number of special subsegments.

Input

The first line contains one integer n (3 ≤ n ≤ 2 ⋅ 10^5).

The second line contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n). All these integers are pairwise distinct.

Output

Print the number of special subsegments of the given permutation.

Examples

Input


5
3 4 1 5 2


Output


2


Input


3
1 3 2


Output


1

Note

Special subsegments in the first example are [1, 5] and [1, 3].

The only special subsegment in the second example is [1, 3].
#include <bits/stdc++.h>
using namespace std;
const int MAX = 2e5 + 5;
int n, ans, a[MAX], cnt[MAX], lft[MAX], rgt[MAX];
void f(int l, int r, int flag) {
  if (l >= r) return;
  int mid = (l + r - flag) / 2;
  lft[mid + 1] = 0;
  for (int i = mid; i >= l; i--) lft[i] = max(a[i], lft[i + 1]);
  rgt[mid] = 0;
  for (int i = mid + 1; i <= r; i++) rgt[i] = max(a[i], rgt[i - 1]);
  for (int i = mid, j = mid + 1; i >= l; i--) {
    while (j <= r && rgt[j] <= lft[i]) {
      cnt[a[j]]++;
      j++;
    }
    if (lft[i] > a[i]) ans += cnt[lft[i] - a[i]];
  }
  for (int i = l; i <= r; i++) cnt[a[i]] = 0;
  f(l, mid, flag);
  f(mid + 1, r, flag);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  f(1, n, 0);
  reverse(a + 1, a + n + 1);
  f(1, n, 1);
  printf("%d\n", ans);
  return 0;
}
