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 loc[200005];
int p[200005];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
    loc[p[i]] = i;
  }
  set<int> s;
  s.insert(0);
  s.insert(n + 1);
  int ans = 0;
  for (int i = n; i >= 1; i--) {
    auto it = s.upper_bound(loc[i]);
    int r = *it - 1;
    it--;
    int l = *it + 1;
    if (r - loc[i] < loc[i] - l) {
      for (int j = loc[i] + 1; j <= r; j++) {
        int need = i - p[j];
        if (need > 0 && loc[need] < loc[i] && loc[need] >= l) ans++;
      }
    } else {
      for (int j = l; j < loc[i]; j++) {
        int need = i - p[j];
        if (need > 0 && loc[need] > loc[i] && loc[need] <= r) ans++;
      }
    }
    s.insert(loc[i]);
  }
  printf("%d\n", ans);
}
