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 + 5;
int a[maxn], le[maxn], ri[maxn];
int n;
stack<int> s;
map<int, int> mm;
void cleas() {
  while (!s.empty()) {
    s.pop();
  }
}
long long mono() {
  long long res = 0;
  int i = 1, j, k;
  cleas();
  a[0] = a[n + 1] = 0x3f3f3f3f;
  while (i <= n + 1) {
    while (!s.empty() && a[s.top()] < a[i]) {
      ri[s.top()] = i;
      s.pop();
    }
    s.push(i++);
  }
  cleas();
  i = n;
  while (i >= 0) {
    while (!s.empty() && a[s.top()] < a[i]) {
      le[s.top()] = i;
      s.pop();
    }
    s.push(i--);
  }
  for (i = 1; i <= n; i++) {
    if (ri[i] - i < i - le[i]) {
      for (j = i + 1; j <= ri[i] - 1; j++) {
        k = a[i] - a[j];
        if (mm[k] > le[i] && mm[k] < i) {
          res++;
        }
      }
    } else {
      for (j = le[i] + 1; j <= i - 1; j++) {
        k = a[i] - a[j];
        if (mm[k] > i && mm[k] < ri[i]) {
          res++;
        }
      }
    }
  }
  return res;
}
int main(int argc, char const *argv[]) {
  int i;
  long long ans;
  mm.clear();
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    mm[a[i]] = i;
  }
  ans = mono();
  printf("%lld\n", ans);
  return 0;
}
