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 N = 200043;
int lf[N];
int rg[N];
int n;
int ans = 0;
int p[N];
int q[N];
void update(int l, int r, int l2, int r2, int sum) {
  for (int i = l; i <= r; i++) {
    int o = sum - p[i];
    if (o >= 1 && o <= n && l2 <= q[o] && q[o] <= r2) ans++;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &p[i]);
    q[p[i]] = i;
  }
  stack<pair<int, int> > s;
  s.push(make_pair(n + 1, -1));
  for (int i = 0; i < n; i++) {
    while (s.top().first < p[i]) s.pop();
    lf[i] = s.top().second;
    s.push(make_pair(p[i], i));
  }
  while (!s.empty()) s.pop();
  s.push(make_pair(n + 1, n));
  for (int i = n - 1; i >= 0; i--) {
    while (s.top().first < p[i]) s.pop();
    rg[i] = s.top().second;
    s.push(make_pair(p[i], i));
  }
  for (int i = 0; i < n; i++) {
    int lenl = i - lf[i] - 1;
    int lenr = rg[i] - i - 1;
    if (lenl == 0 || lenr == 0) continue;
    update(lf[i] + 1, i - 1, i + 1, rg[i] - 1, p[i]);
  }
  cout << ans << endl;
}
