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;
int pos[N], a[N], st[N][20], n;
long long ans = 0;
void init() {
  for (int i = 1; i <= n; i++) st[i][0] = a[i];
  for (int i = 1; i < 20; i++) {
    for (int j = 1; j + (1 << i) - 1 <= n; j++) {
      st[j][i] = max(st[j][i - 1], st[j + (1 << (i - 1))][i - 1]);
    }
  }
}
int q(int l, int r) {
  int k = log2(r - l + 1);
  return max(st[l][k], st[r - (1 << k) + 1][k]);
}
int ef(int l, int r, int x) {
  int mid = (r + l) / 2;
  if (l == r) return l;
  if (q(l, mid) == x) return ef(l, mid, x);
  return ef(mid + 1, r, x);
}
void count(int l, int r) {
  if (l >= r) return;
  int mid = (l + r) / 2, p = 0, Max = 0;
  Max = q(l, r);
  p = ef(l, r, Max);
  if (p < mid) {
    for (int i = l; i < p; i++) {
      int t = a[i];
      if (pos[Max - t] > p && pos[Max - t] <= r) ans++;
    }
  } else {
    for (int i = p + 1; i <= r; i++) {
      int t = a[i];
      if (pos[Max - t] < p && pos[Max - t] >= l) ans++;
    }
  }
  count(l, p - 1);
  count(p + 1, r);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
  }
  init();
  count(1, n);
  printf("%lld", ans);
  return 0;
}
