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 = 2e5 + 10;
const int inf = (int)1e9 + 1;
int a[N], pos[N + 1];
int main() {
  int n;
  while (scanf("%d", &n) != EOF) {
    for (int i = 0; i < n; i++) {
      scanf("%d", &a[i]);
      pos[a[i]] = i;
    }
    int pr[n], ne[n];
    vector<pair<int, int> > st;
    st.push_back({inf, -1});
    for (int i = 0; i < n; i++) {
      while (a[i] >= st.back().first) {
        st.pop_back();
      }
      pr[i] = st.back().second + 1;
      st.push_back({a[i], i});
    }
    st.clear();
    st.push_back({inf, n});
    for (int i = n - 1; i > -1; i--) {
      while (a[i] >= st.back().first) {
        st.pop_back();
      }
      ne[i] = st.back().second - 1;
      st.push_back({a[i], i});
    }
    int ans = 0;
    for (int i = 0; i < n; i++) {
      if (i - pr[i] < ne[i] - i) {
        for (int j = pr[i]; j < i + 1; j++) {
          int need = a[i] - a[j];
          if (need >= 1 && need <= n && pos[need] >= i && pos[need] <= ne[i]) {
            ans++;
          }
        }
      } else {
        for (int j = i; j < ne[i] + 1; j++) {
          int need = a[i] - a[j];
          if (need >= 1 && need <= n && pos[need] >= pr[i] && pos[need] <= i) {
            ans++;
          }
        }
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}
