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 n, a[200005], l[200005], r[200005], p[200005], res;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  stack<int> s;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    p[a[i]] = i;
    while (!s.empty() && a[s.top()] < a[i]) r[s.top()] = i - 1, s.pop();
    l[i] = (s.empty() ? 1 : s.top() + 1);
    s.push(i);
  }
  while (!s.empty()) r[s.top()] = n, s.pop();
  for (int i = 1; i <= n; i++)
    if (l[i] != i && r[i] != i && r[i] - l[i] > 1) {
      if (r[i] - i < i - l[i]) {
        for (int j = i + 1; j <= r[i]; j++) {
          int pos = p[a[i] - a[j]];
          if (l[i] <= pos && pos < i) res++;
        }
      } else {
        for (int j = l[i]; j < i; j++) {
          int pos = p[a[i] - a[j]];
          if (i < pos && pos <= r[i]) res++;
        }
      }
    }
  cout << res;
}
