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 + 5;
int n, a[N], id[N], mx[N];
int solve(int L, int R) {
  if (R - L == 1) return 0;
  int mid = L + R >> 1, ans = solve(L, mid) + solve(mid, R);
  mx[mid - 1] = a[mid - 1], mx[mid] = a[mid];
  for (int i = mid + 1; i < R; i++) mx[i] = max(mx[i - 1], a[i]);
  for (int i = mid - 2; i >= L; i--) mx[i] = max(mx[i + 1], a[i]);
  for (int i = L, j = id[mx[i] - a[i]]; i < R; ++i, j = id[mx[i] - a[i]])
    ans += (L <= min(i, j) && min(i, j) < mid && mid <= max(i, j) &&
            max(i, j) < R && mx[i] > mx[j]);
  return ans;
}
int main() {
  cin >> n, id[0] = n;
  for (int i = 0; i < n; i++) cin >> a[i], id[a[i]] = i;
  cout << solve(0, n);
}
