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>
int q[200005], be = 1, en, n, p[200005], pos[200005], L[200005], R[200005];
long long ans;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]), pos[p[i]] = i;
  p[n + 1] = 200005;
  p[0] = 200005;
  for (int i = 0; i <= n + 1; i++) {
    for (; be <= en && p[q[en]] < p[i]; en--)
      ;
    L[i] = q[en];
    q[++en] = i;
  }
  for (int i = n + 1; i >= 0; i--) {
    for (; be <= en && p[q[en]] < p[i]; en--)
      ;
    R[i] = q[en];
    q[++en] = i;
  }
  for (int i = 1; i <= n; i++) {
    if (i - L[i] < R[i] - i)
      for (int j = L[i] + 1; j < i; j++) {
        if (R[i] > pos[p[i] - p[j]] && pos[p[i] - p[j]] > i) ans++;
      }
    else
      for (int j = R[i] - 1; i < j; j--) {
        if (L[i] < pos[p[i] - p[j]] && pos[p[i] - p[j]] < i) ans++;
      }
  }
  printf("%I64d\n", ans);
}
