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 neg = 1, num = 0;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') neg = -1;
  for (; isdigit(c); c = getchar()) num = (num << 1) + (num << 3) + c - '0';
  return neg * num;
}
int n, p[200010], top, s[200010], a[200010], L[200010], R[200010];
long long ans;
int main() {
  n = read();
  for (int i = 1; i <= n; i++)
    p[i] = read(), R[i] = n + 1, L[i] = 0, a[p[i]] = i;
  for (int i = 1; i <= n; L[i] = s[top], s[++top] = i, i++)
    for (; top && p[i] > p[s[top]];) R[s[top--]] = i;
  for (int i = 1; i <= n; i++) {
    if (i - L[i] + 1 <= R[i] - i + 1) {
      for (int j = max(1, L[i]); j < i; j++) {
        int x = p[i] - p[j];
        if (x <= 0) continue;
        if (a[x] > i && a[x] <= R[i]) ans++;
      }
    } else {
      for (int j = i + 1; j <= min(n, R[i]); j++) {
        int x = p[i] - p[j];
        if (x <= 0) continue;
        if (a[x] < i && a[x] >= L[i]) ans++;
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}
