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 maxn = 2e5 + 10;
int n;
int a[maxn];
int pos[maxn];
int l[maxn], r[maxn];
void calc_l(void) {
  stack<pair<int, int> > stk;
  for (int i = 1; i <= n; i++) {
    while (stk.size() && stk.top().first < a[i]) stk.pop();
    if (stk.size())
      l[i] = stk.top().second + 1;
    else
      l[i] = 1;
    stk.push({a[i], i});
  }
}
void calc_r(void) {
  stack<pair<int, int> > stk;
  for (int i = n; i >= 1; i--) {
    while (stk.size() && stk.top().first < a[i]) stk.pop();
    if (stk.size())
      r[i] = stk.top().second - 1;
    else
      r[i] = n;
    stk.push({a[i], i});
  }
}
int main(void) {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
  }
  calc_l();
  calc_r();
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    if (i - l[i] <= r[i] - i) {
      for (int j = l[i]; j < i; j++)
        if (pos[a[i] - a[j]] > i && pos[a[i] - a[j]] <= r[i]) ans++;
    } else {
      for (int j = i + 1; j <= r[i]; j++)
        if (pos[a[i] - a[j]] < i && pos[a[i] - a[j]] >= l[i]) ans++;
    }
  }
  printf("%lld\n", ans);
}
