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;
int a[200005], pos[200005], pnt[200005], siz[200005];
set<int> s[200005];
int read() {
  char c = getchar();
  int x = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
int fa(int x) {
  if (pnt[x] == x)
    return x;
  else
    return pnt[x] = fa(pnt[x]);
}
void merge(int x, int y) {
  int fx = fa(x), fy = fa(y);
  if (siz[fx] > siz[fy]) swap(fx, fy);
  pnt[fx] = fy;
  for (set<int>::iterator it = s[fx].begin(); it != s[fx].end(); ++it)
    s[fy].insert(*it);
  siz[fy] += siz[fx];
}
int main() {
  int n = read();
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    pos[a[i]] = i;
    s[i].insert(a[i]);
    pnt[i] = i;
    siz[i] = 1;
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    int j = pos[i];
    if (j > 1 && j < n && a[j - 1] < a[j] && a[j + 1] < a[j]) {
      int fx = fa(j - 1), fy = fa(j + 1);
      if (siz[fx] > siz[fy]) swap(fx, fy);
      for (set<int>::iterator it = s[fx].begin(); it != s[fx].end(); ++it) {
        if (s[fy].find(i - (*it)) != s[fy].end()) {
          ++ans;
        }
      }
    }
    if (j > 1 && a[j - 1] < a[j]) merge(j - 1, j);
    if (j < n && a[j + 1] < a[j]) merge(j + 1, j);
  }
  printf("%d\n", ans);
  return 0;
}
