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 main() {
  int n;
  cin >> n;
  int a[n];
  map<int, int> ps;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    ps[a[i]] = i;
  }
  stack<pair<int, int>> st;
  st.push({1e9, n});
  vector<int> rg(n);
  for (int i = n - 1; i >= 0; i--) {
    while (st.top().first < a[i]) {
      st.pop();
    }
    rg[i] = st.top().second;
    st.push({a[i], i});
  }
  st = stack<pair<int, int>>();
  st.push({1e9, -1});
  int ans = 0;
  for (int i = 0; i < n; i++) {
    while (st.top().first < a[i]) {
      st.pop();
    }
    int l = st.top().second;
    int r = rg[i];
    st.push({a[i], i});
    if (i - l > r - i) {
      for (int o = i + 1; o < r; o++) {
        if (ps[a[i] - a[o]] > l && ps[a[i] - a[o]] < i) {
          ans++;
        }
      }
    } else {
      for (int o = i - 1; o > l; o--) {
        if (ps[a[i] - a[o]] < r && ps[a[i] - a[o]] > i) {
          ans++;
        }
      }
    }
  }
  cout << ans << endl;
}
