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, ans;
int a[300005];
int pos[300005];
int l[300005];
int r[300005];
void doStackTrick() {
  stack<pair<int, int> > s;
  for (int i = N; i >= 1; i--) {
    while (!s.empty() && s.top().first < a[i]) s.pop();
    if (s.empty())
      r[i] = N;
    else
      r[i] = s.top().second - 1;
    s.push({a[i], i});
  }
  while (!s.empty()) s.pop();
  for (int i = 1; i <= N; i++) {
    while (!s.empty() && s.top().first < a[i]) s.pop();
    if (s.empty())
      l[i] = 1;
    else
      l[i] = s.top().second + 1;
    s.push({a[i], i});
  }
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> N;
  for (int i = 1; i <= N; i++) {
    cin >> a[i];
    pos[a[i]] = i;
  }
  doStackTrick();
  for (int i = 2; i < N; i++)
    if (l[i] < i && r[i] > i) {
      if (i - l[i] < r[i] - i) {
        for (int j = i - 1; j >= l[i]; j--) {
          int p = pos[a[i] - a[j]];
          if (p > i && p <= r[i]) ans++;
        }
      } else {
        for (int j = i + 1; j <= r[i]; j++) {
          int p = pos[a[i] - a[j]];
          if (p < i && p >= l[i]) ans++;
        }
      }
    }
  cout << ans;
  return 0;
}
