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[200005], l[200005], r[200005], s[200005], cnt, s2[200005], b[200005],
    pos[200005];
int main(int argc, char** argv) {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
  }
  for (int i = 1; i <= n; i++) {
    while (cnt && s[cnt] < a[i]) --cnt;
    l[i] = s2[cnt];
    s[++cnt] = a[i];
    s2[cnt] = i;
  }
  cnt = 0;
  for (int i = n; i >= 1; i--) {
    while (cnt && s[cnt] < a[i]) --cnt;
    r[i] = s2[cnt];
    s[++cnt] = a[i];
    s2[cnt] = i;
    if (!r[i]) r[i] = n + 1;
    ++l[i], --r[i];
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    if (i - l[i] < r[i] - i) {
      for (int j = i; j >= l[i]; j--)
        ans += (pos[a[i] - a[j]] <= r[i] && pos[a[i] - a[j]] > i);
    } else {
      for (int j = r[i]; j >= i; j--)
        ans += (pos[a[i] - a[j]] <= i && pos[a[i] - a[j]] >= l[i]);
    }
  }
  cout << ans;
  return 0;
}
