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;
const int N = 2e5 + 5;
int n, a[N], l[N], r[N], k, ans, pos[N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i), pos[a[i]] = i;
  for (int i = 1; i <= n; ++i)
    for (l[i] = i - 1; l[i] >= 1 && a[i] > a[l[i]]; l[i] = l[l[i]])
      ;
  for (int i = n; i >= 1; --i)
    for (r[i] = i + 1; r[i] <= n && a[i] > a[r[i]]; r[i] = r[r[i]])
      ;
  for (int i = 1; i <= n; ++i)
    if (i - l[i] < r[i] - i)
      for (int j = l[i] + 1; j < i; ++j) {
        k = pos[a[i] - a[j]];
        if (i < k && k < r[i]) ++ans;
      }
    else
      for (int j = r[i] - 1; i < j; --j) {
        k = pos[a[i] - a[j]];
        if (l[i] < k && k < i) ++ans;
      }
  printf("%d", ans);
}
