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 N = 200004;
int n, a[N], l[N], r[N], id[N];
stack<int> st;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    id[a[i]] = i;
  }
  for (int i = 1; i <= n; i++) {
    while (!st.empty() && a[st.top()] <= a[i]) st.pop();
    if (st.empty())
      l[i] = 0;
    else
      l[i] = st.top();
    st.push(i);
  }
  while (!st.empty()) st.pop();
  for (int i = n; i >= 1; i--) {
    while (!st.empty() && a[st.top()] <= a[i]) st.pop();
    if (st.empty())
      r[i] = n + 1;
    else
      r[i] = st.top();
    st.push(i);
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    if (i - l[i] < r[i] - i)
      for (int j = l[i] + 1; j < i; j++) {
        if (id[a[i] - a[j]] > i && id[a[i] - a[j]] < r[i]) ans++;
      }
    else
      for (int j = r[i] - 1; j > i; j--) {
        if (id[a[i] - a[j]] < i && id[a[i] - a[j]] > l[i]) ans++;
      }
  }
  cout << ans;
}
