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 = 998244353;
int n, a[200005], t, pos[200005], ans;
struct node {
  int l, r;
} q[200005], now;
void solve(int l, int mid, int r) {
  if (mid - l < r - mid) {
    for (int i = l; i < mid; i++)
      if (mid < pos[a[mid] - a[i]] && pos[a[mid] - a[i]] <= r) ans++;
  } else {
    for (int i = mid + 1; i <= r; i++)
      if (l <= pos[a[mid] - a[i]] && pos[a[mid] - a[i]] < mid) ans++;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), pos[a[i]] = i;
  a[n + 1] = 1e9;
  for (int i = 1; i <= n + 1; i++) {
    now.l = now.r = i;
    while (t && a[i] > a[q[t].r])
      solve(q[t].l, q[t].r, i - 1), now.l = q[t].l, t--;
    q[++t] = now;
  }
  printf("%d\n", ans);
}
