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;
long long n;
long long p[200011];
long long pos[200011];
long long l[200011];
long long r[200011];
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> p[i];
    pos[p[i]] = i;
  }
  p[0] = p[n + 1] = 200001;
  stack<long long> CurrList;
  CurrList.push(0);
  for (long long i = 1; i <= n; i++) {
    while (!CurrList.empty() && p[CurrList.top()] < p[i]) CurrList.pop();
    l[p[i]] = CurrList.top();
    CurrList.push(i);
  }
  while (!CurrList.empty()) CurrList.pop();
  CurrList.push(n + 1);
  for (long long i = n; i >= 1; i--) {
    while (!CurrList.empty() && p[CurrList.top()] < p[i]) CurrList.pop();
    r[p[i]] = CurrList.top();
    CurrList.push(i);
  }
  long long res = 0;
  for (long long i = 2; i <= n; i++) {
    if (pos[i] - l[i] < r[i] - pos[i]) {
      for (long long j = pos[i] - 1; j > l[i]; j--) {
        long long pos_x = pos[i - p[j]];
        if (pos[i] < pos_x && pos_x < r[i]) res++;
      }
    } else {
      for (long long j = pos[i] + 1; j < r[i]; j++) {
        long long pos_x = pos[i - p[j]];
        if (l[i] < pos_x && pos_x < pos[i]) res++;
      }
    }
  }
  cout << res;
}
