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 mod = 1e9 + 7;
const int maxn = 1e6 + 10;
int a[maxn];
int id[maxn];
int ans;
int dp[maxn][20];
int query(int l, int r) {
  int k = log2(r - l + 1);
  ;
  return max(dp[l][k], dp[r - (1 << k) + 1][k]);
}
void dfs(int l, int r) {
  if (r - l <= 1) return;
  int mx = id[query(l, r)];
  ;
  if (mx - l > r - mx) {
    for (int i = mx + 1; i <= r; i++) {
      int t = a[i];
      if (id[a[mx] - a[i]] >= l && id[a[mx] - a[i]] < mx) {
        ans++;
      }
    }
  } else {
    for (int i = l; i < mx; i++) {
      int t = a[i];
      if (id[a[mx] - a[i]] > mx && id[a[mx] - a[i]] <= r) {
        ans++;
      }
    }
  }
  dfs(l, mx - 1);
  dfs(mx + 1, r);
}
int main() {
  std::ios::sync_with_stdio(false);
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    id[a[i]] = i;
  }
  for (int i = 1; i <= n; i++) {
    dp[i][0] = a[i];
  }
  for (int j = 1; (1 << j) <= n; j++) {
    for (int i = 1; i + (1 << j) - 1 <= n; i++) {
      dp[i][j] = max(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
    }
  }
  dfs(1, n);
  cout << ans << endl;
}
