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 n;
int ans = 0;
int a[200050], pos[200050];
void solve(int l, int r, int l2, int r2, int cur) {
  for (int i = l; i <= r; i++) {
    int find = cur - a[i];
    if (find >= 1 && find <= n) {
      if (pos[find] >= l2 && pos[find] <= r2) ans++;
    }
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i], pos[a[i]] = i;
  int lm[n + 1], rm[n + 1];
  stack<pair<int, int> > st;
  st.push({n + 1, 0});
  for (int i = 1; i <= n; i++) {
    while (st.top().first < a[i]) st.pop();
    lm[i] = st.top().second;
    st.push({a[i], i});
  }
  while (st.size()) st.pop();
  st.push({n + 1, n + 1});
  for (int i = n; i >= 1; i--) {
    while (st.top().first < a[i]) st.pop();
    rm[i] = st.top().second;
    st.push({a[i], i});
  }
  for (int i = 1; i <= n; i++) {
    int left = i - lm[i] - 1;
    int right = rm[i] - i - 1;
    if (left < right)
      solve(lm[i] + 1, i - 1, i + 1, rm[i] - 1, a[i]);
    else
      solve(i + 1, rm[i] - 1, lm[i] + 1, i - 1, a[i]);
  }
  cout << ans << endl;
}
