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;
int a[200005], b[200005], l[200005], r[200005], pos[200005];
inline int in() {
  int res, ff;
  res = ff = 0;
  char ch;
  if ((ch = getchar()) == '-')
    ff = 1;
  else if (ch >= '0' && ch <= '9')
    res = ch - '0';
  while ((ch = getchar()) >= '0' && ch <= '9') res = res * 10 + (ch - '0');
  return ff ? -res : res;
}
int n;
int main() {
  n = in();
  for (int i = 1; i <= n; i++) {
    a[i] = in();
    pos[a[i]] = i;
    l[i] = r[i] = i;
  }
  a[0] = a[n + 1] = n + 1;
  for (int i = 1; i <= n; i++)
    while (a[l[i] - 1] < a[i]) l[i] = l[l[i] - 1];
  for (int i = n; i >= 1; i--)
    while (a[r[i] + 1] < a[i]) r[i] = r[r[i] + 1];
  for (int i = 1; i <= n; i++) {
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    if (i - l[i] <= r[i] - i)
      for (int j = l[i]; j < i; j++) {
        if (pos[a[i] - a[j]] <= r[i] && pos[a[i] - a[j]] > i) ans++;
      }
    else {
      for (int j = i + 1; j <= r[i]; j++)
        if (pos[a[i] - a[j]] >= l[i] && pos[a[i] - a[j]] < i) ans++;
    }
  }
  printf("%d\n", ans);
}
