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 inf = 0x3f3f3f3f;
const int maxn = 2e5 + 9;
int a[maxn], pt[maxn];
stack<int> st;
int le[maxn], ri[maxn];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    pt[a[i]] = i;
  }
  a[0] = inf;
  st.push(0);
  for (int i = 1; i <= n; i++) {
    while (!st.empty() && a[st.top()] < a[i]) st.pop();
    le[i] = st.top();
    st.push(i);
  }
  while (!st.empty()) st.pop();
  st.push(n + 1);
  a[n + 1] = inf;
  for (int i = n; i >= 1; i--) {
    while (!st.empty() && a[st.top()] < a[i]) st.pop();
    ri[i] = st.top();
    st.push(i);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int L = le[i] + 1;
    int R = ri[i] - 1;
    if (L == R) continue;
    if (i - L <= R - i) {
      for (int j = L; j < i; j++) {
        int val = a[i] - a[j];
        if (pt[val] <= R && pt[val] > i) ans++;
      }
    } else {
      for (int j = i + 1; j <= R; j++) {
        int val = a[i] - a[j];
        if (pt[val] >= L && pt[val] < i) ans++;
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}
