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 + 5;
int n;
int a[N], b[N], mx[N];
long long ans = 0;
inline void solve(int l, int r) {
  if (l == r) return;
  int md = l + r >> 1, i;
  solve(l, md);
  solve(md + 1, r);
  mx[md] = a[md];
  mx[md + 1] = a[md + 1];
  for (i = md - 1; i >= l; --i) mx[i] = max(mx[i + 1], a[i]);
  for (i = md + 2; i <= r; ++i) mx[i] = max(mx[i - 1], a[i]);
  for (i = md; i >= l; --i) {
    int t = b[mx[i] - a[i]];
    if (t > md and t <= r and mx[t] <= mx[i]) ++ans;
  }
  for (int i = md + 1; i <= r; i++) {
    int t = b[mx[i] - a[i]];
    if (t <= md and t >= l and mx[t] <= mx[i]) ++ans;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i], b[a[i]] = i;
  solve(1, n);
  cout << ans << endl;
  return 0;
}
