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 main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  vector<int> p(n);
  for (int& e : p) cin >> e;
  vector<int> pos(n + 1);
  for (int i = 0; i < n; ++i) {
    pos[p[i]] = i;
  }
  vector<int> s, l(n, -1), r(n, n);
  for (int i = 0; i < n; ++i) {
    while (!s.empty() && p[s.back()] < p[i]) s.pop_back();
    if (!s.empty()) l[i] = s.back();
    s.push_back(i);
  }
  s.clear();
  for (int i = n - 1; i >= 0; --i) {
    while (!s.empty() && p[s.back()] < p[i]) s.pop_back();
    if (!s.empty()) r[i] = s.back();
    s.push_back(i);
  }
  int cnt = 0;
  for (int i = 0; i < n; ++i) {
    int from, to;
    if (i - l[i] < r[i] - i) {
      from = l[i] + 1;
      to = i;
    } else {
      from = i + 1;
      to = r[i];
    }
    for (int j = from; j < to; ++j) {
      int ps = pos[p[i] - p[j]];
      if (ps <= l[i] || ps >= r[i]) continue;
      int val = (j < i ? +1 : -1) * (ps < i ? +1 : -1);
      if (val < 0) ++cnt;
    }
  }
  cout << cnt << '\n';
}
