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>
const int N = 200005;
int n, p[N], pp[N], l[N], r[N], st[N], top, ans;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i)
    scanf("%d", p + i), pp[p[i]] = i, l[i] = 0, r[i] = n + 1;
  for (int i = 1; i <= n; ++i) {
    while (top > 0 && p[i] > p[st[top]]) r[st[top--]] = i;
    st[++top] = i;
  }
  top = 0;
  for (int i = n; i >= 1; --i) {
    while (top > 0 && p[i] > p[st[top]]) l[st[top--]] = i;
    st[++top] = 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 (pp[p[i] - p[j]] > i && pp[p[i] - p[j]] < r[i]) ++ans;
    } else {
      for (int j = i + 1; j < r[i]; ++j)
        if (pp[p[i] - p[j]] < i && pp[p[i] - p[j]] > l[i]) ++ans;
    }
  }
  printf("%d\n", ans);
  return 0;
}
