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;
inline int read() {
  int s = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') s = s * 10 + ch - '0', ch = getchar();
  return s * w;
}
const int sea = 1e6 + 7;
int n, ans, a[sea], l[sea], r[sea], st[sea], id[sea], top;
int main() {
  n = read();
  for (int i = 1; i <= n; i++) a[i] = read(), id[a[i]] = i;
  top = 0;
  for (int i = 1; i <= n; i++) {
    while (top && a[st[top]] < a[i]) top--;
    l[i] = st[top];
    st[++top] = i;
  }
  top = 0;
  st[top] = n + 1;
  for (int i = n; i >= 1; i--) {
    while (top && a[st[top]] < a[i]) top--;
    r[i] = st[top];
    st[++top] = i;
  }
  for (int i = 1; i <= n; i++) {
    if (i - l[i] < r[i] - i) {
      for (int j = l[i] + 1; j <= i; j++)
        if (id[a[i] - a[j]] > i && id[a[i] - a[j]] < r[i]) ++ans;
    } else {
      for (int j = i + 1; j <= r[i] - 1; j++)
        if (id[a[i] - a[j]] < i && id[a[i] - a[j]] > l[i]) ++ans;
    }
  }
  printf("%d\n", ans);
  return 0;
}
