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 double eps = 1e-6;
const int mod = 1e9 + 7;
const int maxn = 2e6 + 11;
const int maxm = 1e5 + 100;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
int a[maxn];
int l[maxn], r[maxn];
int idx[maxn];
long long ans = 0;
void gao(int x) {
  int L, R;
  int pL, pR;
  if (x - l[x] < r[x] - x) {
    L = l[x];
    R = x - 1;
    pL = x + 1;
    pR = r[x];
  } else {
    pL = l[x];
    pR = x - 1;
    L = x + 1;
    R = r[x];
  }
  for (int i = L; i <= R; i++) {
    int y = a[x] - a[i];
    if (idx[y] >= pL && idx[y] <= pR) ans++;
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    idx[a[i]] = i;
    l[i] = r[i] = i;
  }
  a[0] = a[n + 1] = n + 1;
  for (int i = 1; i <= n; i++) {
    while (a[l[i] - 1] <= a[i]) {
      l[i] = l[l[i] - 1];
    }
  }
  for (int i = n; i >= 1; i--) {
    while (a[r[i] + 1] <= a[i]) {
      r[i] = r[r[i] + 1];
    }
  }
  ans = 0;
  for (int i = 1; i <= n; i++) {
    gao(i);
  }
  printf("%lld", ans);
  return 0;
}
