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 maxn = 2e5 + 10, mod = 1e9 + 7, INF = 0x3f3f3f3f;
int a[maxn];
long long ans = 0;
set<int> st;
void solve(int l, int r) {
  if (l == r) return;
  int m = (l + r) >> 1, mx = 0;
  st.clear();
  for (int i = m, j = m + 1; i >= l; --i) {
    mx = max(mx, a[i]);
    while (j <= r && a[j] < mx) {
      st.insert(a[j]);
      ++j;
    }
    ans += st.count(mx - a[i]);
  }
  mx = 0;
  st.clear();
  for (int i = m, j = m + 1; j <= r; ++j) {
    mx = max(mx, a[j]);
    while (i >= l && a[i] < mx) {
      st.insert(a[i]);
      --i;
    }
    ans += st.count(mx - a[j]);
  }
  solve(l, m);
  solve(m + 1, r);
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  solve(1, n);
  cout << ans;
  return 0;
}
