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 inf = 0x3f3f3f3f;
const double pi = 3.1415926535897932384626;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
const int maxn = 2e5 + 5;
long long n, ans, a[maxn];
set<long long> st;
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  for (int i = 2; i < n; ++i) {
    st.clear();
    if (a[i] < a[i - 1] || a[i] < a[i + 1]) continue;
    for (int j = i - 1; j >= 1 && a[j] < a[i]; --j) st.insert(a[j]);
    for (int j = i + 1; j <= n && a[j] < a[i]; ++j) {
      if (st.count(a[i] - a[j])) ++ans;
    }
  }
  printf("%lld\n", ans);
  return 0;
}
