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;
using INT = long long;
using pii = pair<int, int>;
int p[300100], id[300100], pre[300100], nxt[300100];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
    id[p[i]] = i;
  }
  pre[1] = 1;
  for (int i = 2; i <= n; i++) {
    int pos = i - 1;
    while (pos && p[pos] < p[i]) pos = pre[pos] - 1;
    pre[i] = pos + 1;
  }
  nxt[n] = n;
  for (int i = n - 1; i >= 1; i--) {
    int pos = i + 1;
    while (pos <= n && p[pos] < p[i]) pos = nxt[pos] + 1;
    nxt[i] = pos - 1;
  }
  INT ans = 0;
  for (int i = 1; i <= n; i++) {
    if (pre[i] == i or nxt[i] == i) continue;
    if (i - pre[i] < nxt[i] - i) {
      for (int j = pre[i]; j < i; j++)
        if (id[p[i] - p[j]] > i && id[p[i] - p[j]] <= nxt[i]) ans++;
    } else {
      for (int j = i + 1; j <= nxt[i]; j++)
        if (id[p[i] - p[j]] < i && id[p[i] - p[j]] >= pre[i]) ans++;
    }
  }
  cout << ans << endl;
  return 0;
}
