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 VI = vector<int>;
using VLL = vector<LL>;
const int __n = 2e5 + 5, Mod = 1e9 + 7, inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
int n, a[__n], p[__n], lft[__n], rft[__n];
void __run() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i], p[a[i]] = i;
  stack<int> sta;
  a[0] = a[n + 1] = n + 1;
  for (int i = 1; i <= n + 1; i++) {
    while (!sta.empty() && sta.top() < a[i])
      rft[p[sta.top()]] = i - 1, sta.pop();
    sta.push(a[i]);
  }
  while (!sta.empty()) sta.pop();
  for (int i = n; i >= 0; i--) {
    while (!sta.empty() && sta.top() < a[i])
      lft[p[sta.top()]] = i + 1, sta.pop();
    sta.push(a[i]);
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    if (i - lft[i] < rft[i] - i)
      for (int j = lft[i]; j < i; j++)
        if (p[a[i] - a[j]] > i && p[a[i] - a[j]] <= rft[i]) ans++;
    if (i - lft[i] >= rft[i] - i)
      for (int j = i + 1; j <= rft[i]; j++)
        if (p[a[i] - a[j]] >= lft[i] && p[a[i] - a[j]] < i) ans++;
  }
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  __run();
  return 0;
}
