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 LL = long long;
const int maxn = 2e5 + 5;
int pos[maxn];
int a[maxn];
int le[maxn], ri[maxn];
stack<int> sta;
int main() {
  int n, i;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
  }
  sta.push(1);
  for (i = 2; i <= n; i++) {
    while (sta.size() && a[sta.top()] < a[i]) {
      ri[sta.top()] = i;
      sta.pop();
    }
    sta.push(i);
  }
  while (sta.size()) {
    ri[sta.top()] = n + 1;
    sta.pop();
  }
  sta.push(n);
  for (i = n - 1; i > 0; i--) {
    while (sta.size() && a[sta.top()] < a[i]) {
      le[sta.top()] = i;
      sta.pop();
    }
    sta.push(i);
  }
  while (sta.size()) sta.pop();
  int ln, rn, need;
  int ans = 0;
  int j;
  for (i = 1; i <= n; i++) {
    ln = i - le[i] - 1;
    rn = ri[i] - i - 1;
    if (ln == 0 || rn == 0) continue;
    if (ln < rn) {
      for (j = le[i] + 1; j < i; j++) {
        need = a[i] - a[j];
        if (pos[need] < ri[i] && pos[need] > i) ans++;
      }
    } else {
      for (j = i + 1; j < ri[i]; j++) {
        need = a[i] - a[j];
        if (pos[need] > le[i] && pos[need] < i) ans++;
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}
