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;
void solve() {
  int n;
  cin >> n;
  vector<int> arr(n);
  vector<pair<int, int>> sorted_arr(n);
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    sorted_arr[i] = {arr[i], i};
  }
  sort(sorted_arr.begin(), sorted_arr.end());
  vector<int> pos(n);
  for (int i = 0; i < n; i++) {
    pos[i] = sorted_arr[i].second;
  }
  vector<int> next(n), prev(n);
  next[n - 1] = n;
  for (int i = n - 2; i >= 0; i--) {
    int j = i + 1;
    while (j < n && arr[i] > arr[j]) {
      j = next[j];
    }
    next[i] = j;
  }
  prev[0] = -1;
  for (int i = 1; i < n; i++) {
    int j = i - 1;
    while (j >= 0 && arr[i] > arr[j]) {
      j = prev[j];
    }
    prev[i] = j;
  }
  int ans = 0;
  for (int i = 1; i < n - 1; i++) {
    if (i - prev[i] < next[i] - i) {
      for (int j = prev[i] + 1; j < i; j++) {
        int k = pos[arr[i] - arr[j] - 1];
        if (i < k && k < next[i]) ans++;
      }
    } else {
      for (int j = next[i] - 1; j > i; j--) {
        int k = pos[arr[i] - arr[j] - 1];
        if (prev[i] < k && k < i) ans++;
      }
    }
  }
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
  solve();
  cout << '\n';
}
