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 + 50;
int p[N], rp[N];
int l[N], r[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    rp[p[i]] = i;
  }
  set<int> pos;
  for (int i = n; i > 0; i--) {
    set<int>::iterator it = pos.lower_bound(rp[i]);
    if (it == pos.end())
      r[i] = n;
    else
      r[i] = (*it) - 1;
    if (it == pos.begin())
      l[i] = 1;
    else
      l[i] = *(--it) + 1;
    pos.insert(rp[i]);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int x = rp[p[i]];
    int xl = l[p[i]];
    int xr = r[p[i]];
    if (x - xl < xr - x) {
      for (int j = xl; j < x; j++) {
        int y = p[i] - p[j];
        if (rp[y] > i && rp[y] <= xr) {
          ans++;
        }
      }
    } else {
      for (int j = x + 1; j <= xr; j++) {
        int y = p[i] - p[j];
        if (rp[y] >= xl && rp[y] < i) {
          ans++;
        }
      }
    }
  }
  cout << ans;
}
