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;
using ll = long long;
using ull = unsigned long long;
const ll INF = 1e9;
void solve() {
  int n;
  cin >> n;
  vector<int> p(n + 1);
  vector<int> q(n + 1);
  for (int i = 1; i <= n; ++i) {
    cin >> p[i];
    q[p[i]] = i;
  }
  stack<pair<int, int>> s;
  vector<int> left(n + 1), right(n + 1);
  s.push({INF, 0});
  for (int i = 1; i <= n; ++i) {
    while (s.top().first < p[i]) s.pop();
    left[p[i]] = s.top().second;
    s.push({p[i], i});
  }
  while (!s.empty()) s.pop();
  s.push({INF, n + 1});
  for (int i = n; i >= 1; --i) {
    while (s.top().first < p[i]) s.pop();
    right[p[i]] = s.top().second;
    s.push({p[i], i});
  }
  ll ans = 0;
  for (int x = 1; x <= n; ++x) {
    int l = left[x], r = right[x];
    if (q[x] - l < r - q[x]) {
      for (int i = l + 1; i < q[x]; ++i) {
        int a = x - p[i];
        if (a >= 1 && a <= n && q[a] > q[x] && q[a] < r) ++ans;
      }
    } else {
      for (int i = q[x] + 1; i < r; ++i) {
        int a = x - p[i];
        if (a >= 1 && a <= n && q[a] > l && q[a] < q[x]) ++ans;
      }
    }
  }
  cout << ans;
}
int main() { solve(); }
