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 MX = 200100;
int p[MX], l[MX], r[MX], pos[MX];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int n;
  cin >> n;
  vector<int> was;
  was.reserve(n);
  for (int i(0); i < n; ++i) {
    cin >> p[i];
    pos[p[i]] = i;
    while (!was.empty() and p[was.back()] < p[i]) was.pop_back();
    l[i] = was.empty() ? 0 : was.back() + 1;
    was.push_back(i);
  }
  was.clear();
  for (int i(n - 1); i >= 0; --i) {
    while (!was.empty() and p[was.back()] < p[i]) was.pop_back();
    r[i] = was.empty() ? n - 1 : was.back() - 1;
    was.push_back(i);
  }
  int64_t ans(0);
  for (int v(1); v <= n; ++v) {
    int i(pos[v]);
    if (i - l[i] < r[i] - i) {
      for (int j(l[i]); j < i; ++j) {
        int u(v - p[j]);
        ans += pos[u] > i and pos[u] <= r[i];
      }
    } else {
      for (int j(r[i]); j > i; --j) {
        int u(v - p[j]);
        ans += pos[u] < i and pos[u] >= l[i];
      }
    }
  }
  cout << ans << '\n';
}
