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 res, v[200010], pos[200010], pref[200010], suf[200010];
void upt(int l1, int r1, int l2, int r2, int x) {
  for (int i = l1; i <= r1; i++) {
    int comp = x - v[i];
    if (comp >= 1 && l2 <= pos[comp] && pos[comp] <= r2) res++;
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &v[i]);
    pos[v[i]] = i;
  }
  stack<pair<int, int> > p;
  p.push({n + 1, -1});
  for (int i = 0; i < n; i++) {
    while (p.top().first < v[i]) p.pop();
    pref[i] = p.top().second;
    p.push({v[i], i});
  }
  while (p.size()) p.pop();
  p.push({n + 1, n});
  for (int i = n - 1; i >= 0; i--) {
    while (p.top().first < v[i]) p.pop();
    suf[i] = p.top().second;
    p.push({v[i], i});
  }
  for (int i = 0; i < n; i++) {
    int ltam = i - pref[i] - 1;
    int rtam = suf[i] - i - 1;
    if (!ltam || !rtam) continue;
    if (ltam < rtam)
      upt(pref[i] + 1, i - 1, i + 1, suf[i] - 1, v[i]);
    else
      upt(pref[i] + 1, i - 1, i + 1, suf[i] - 1, v[i]);
  }
  printf("%d\n", res);
  return 0;
}
