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;
inline int read() {
  int x = 0;
  char ch = getchar();
  while (!isdigit(ch)) ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x;
}
int a[200010], pos[200010], L[200010], R[200010], s[200010];
int main() {
  int n = read(), top = 0, ans = 0;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), pos[a[i]] = i;
  for (int i = 1; i <= n; i++) {
    while (top && a[s[top]] < a[i]) top--;
    L[i] = s[top], s[++top] = i;
  }
  s[top = 0] = n + 1;
  for (int i = n; i >= 1; i--) {
    while (top && a[s[top]] < a[i]) top--;
    R[i] = s[top], s[++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 (pos[a[i] - a[j]] > i && pos[a[i] - a[j]] < R[i]) ans++;
    } else {
      for (int j = i + 1; j < R[i]; j++)
        if (pos[a[i] - a[j]] < i && pos[a[i] - a[j]] > L[i]) ans++;
    }
  }
  cout << ans << endl;
  return 0;
}
