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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int maxn = 2e5 + 100;
int p[maxn];
int ja[maxn];
long long maxa[maxn];
long long maxj[maxn];
long long ans, n;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int y = 1; y <= n; y++) cin >> p[y], ja[p[y]] = y;
  p[0] = p[n + 1] = n + 1;
  for (int y = 1; y <= n; y++) {
    int i = y - 1;
    while (p[i] < p[y]) i = maxa[i];
    maxa[y] = i;
  }
  for (int y = n; y; y--) {
    int i = y + 1;
    while (p[i] < p[y]) i = maxj[i];
    maxj[y] = i;
  }
  for (int y = n; y; y--) {
    int i = ja[y];
    if (i - maxa[i] < maxj[i] - i) {
      for (int j = maxa[i] + 1; j < i; j++) {
        ans += (ja[y - p[j]] > i && ja[y - p[j]] < maxj[i]);
      }
    } else {
      for (int j = i + 1; j < maxj[i]; j++)
        ans += (ja[y - p[j]] > maxa[i] && ja[y - p[j]] < i);
    }
  }
  cout << ans;
}
