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 MAX = (int)2e5 + 11;
int n, a[MAX], pos[MAX], lf[MAX], rg[MAX], an;
void calc(int lo, int hi, int pos_lo, int pos_hi, int mx) {
  for (int i = lo; i < hi; ++i) {
    int x = mx - a[i];
    if (x >= 1 && x <= n && pos[x] >= pos_lo && pos[x] < pos_hi) {
      ++an;
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    pos[a[i]] = i;
  }
  vector<pair<int, int>> v;
  v.push_back({MAX, -1});
  for (int i = 0; i < n; ++i) {
    while (v.back().first < a[i]) v.pop_back();
    lf[i] = v.back().second;
    v.push_back({a[i], i});
  }
  v.clear();
  v.push_back({MAX, n});
  for (int i = n - 1; i >= 0; --i) {
    while (v.back().first < a[i]) v.pop_back();
    rg[i] = v.back().second;
    v.push_back({a[i], i});
  }
  for (int i = 0; i < n; ++i) {
    if (i - lf[i] < rg[i] - i) {
      calc(lf[i] + 1, i, i + 1, rg[i], a[i]);
    } else {
      calc(i + 1, rg[i], lf[i] + 1, i, a[i]);
    }
  }
  cout << an << endl;
  return 0;
}
