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>
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
int read() {
  char ch = getchar();
  int x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int N = 2e5 + 5;
int mx[N], a[N], pos[N];
int n;
int ans = 0;
void merge(int l, int r) {
  int mid = (l + r) >> 1;
  if (l + 1 >= r) return;
  merge(mid + 1, r);
  merge(l, mid);
  mx[mid] = a[mid];
  for (int i = mid - 1; i >= l; i--) mx[i] = max(mx[1 + i], a[i]);
  mx[mid + 1] = a[mid + 1];
  for (int i = mid + 2; i <= r; i++) mx[i] = max(mx[i - 1], a[i]);
  for (int i = l; i <= mid - 1; i++) {
    int temp = pos[mx[i] - a[i]];
    if (temp > mid && temp <= r && mx[temp] < mx[i]) {
      ans++;
    }
  }
  for (int i = mid + 2; i <= r; i++) {
    int temp = pos[mx[i] - a[i]];
    if (temp <= mid && temp >= l && mx[temp] < mx[i]) ans++;
  }
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    pos[a[i]] = i;
  }
  merge(1, n);
  cout << ans << endl;
  return 0;
}
