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 = 2e5 + 7;
int n;
vector<int> a, p;
vector<int> e[N];
long long solve(int l, int r) {
  if (l == r) {
    return 0;
  }
  int m = (l + r) / 2;
  long long res = 0;
  vector<int> s;
  int mx_r = 0;
  for (int i = m + 1; i <= r; i++) {
    mx_r = max(mx_r, a[i]);
    s.push_back(mx_r);
    e[mx_r - a[i]].push_back(i);
  }
  int mx_l = 0;
  for (int i = m; i >= l; i--) {
    mx_l = max(mx_l, a[i]);
    int pos = upper_bound(s.begin(), s.end(), mx_l) - s.begin();
    pos += m + 1;
    int need = mx_l - a[i];
    if (m < p[need] && p[need] < pos) {
      res++;
    }
    int ind =
        lower_bound(e[a[i]].begin(), e[a[i]].end(), pos) - e[a[i]].begin();
    res += (int)e[a[i]].size() - ind;
  }
  mx_r = 0;
  for (int i = m + 1; i <= r; i++) {
    mx_r = max(mx_r, a[i]);
    e[mx_r - a[i]].clear();
  }
  return res + solve(l, m) + solve(m + 1, r);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  a.resize(n);
  p.resize(n + 1);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    p[a[i]] = i;
  }
  cout << solve(0, n - 1) << '\n';
  return 0;
}
