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 a[200000];
int logg[200001];
int st[200000][20];
int op[200001];
long long rec(int l, int r) {
  if (l >= r) return 0;
  int j = logg[r - l + 1];
  int maximum = max(st[l][j], st[r - (1 << j) + 1][j]);
  int index = op[maximum];
  long long ans = 0;
  if (index - l < r - index) {
    for (int i = l; i < index; i++) {
      if (op[maximum - a[i]] > index && op[maximum - a[i]] <= r) ans++;
    }
  } else {
    for (int i = r; i > index; i--) {
      if (op[maximum - a[i]] < index && op[maximum - a[i]] >= l) ans++;
    }
  }
  return ans + rec(l, index - 1) + rec(index + 1, r);
}
void solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    op[a[i]] = i;
  }
  logg[1] = 0;
  for (int i = 2; i <= 200000; i++) logg[i] = logg[i / 2] + 1;
  for (int i = 0; i < n; i++) st[i][0] = a[i];
  for (int j = 1; j <= 20; j++)
    for (int i = 0; i + (1 << j) <= n; i++)
      st[i][j] = max(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
  long long ans = rec(0, n - 1);
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}
