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 = 1e6 + 10;
const int inf = INT_MAX;
template <typename T>
void read(T &x) {
  x = 0;
  long long f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x *= f;
}
stack<long long> s;
long long n;
long long a[N], L[N], R[N], pos[N];
signed main() {
  read(n);
  for (long long i = 1; i <= n; i++) read(a[i]), pos[a[i]] = i;
  for (long long i = 1; i <= n; i++) {
    while (!s.empty() && a[s.top()] < a[i]) s.pop();
    if (!s.empty())
      L[i] = s.top();
    else
      L[i] = 0;
    s.push(i);
  }
  while (!s.empty()) s.pop();
  s.push(n + 1);
  for (long long i = n; i >= 1; i--) {
    while (!s.empty() && a[s.top()] < a[i]) s.pop();
    if (!s.empty())
      R[i] = s.top();
    else
      R[i] = n + 1;
    s.push(i);
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    if (i - L[i] < R[i] - i) {
      for (long long j = L[i] + 1; j <= i - 1; j++)
        if (pos[a[i] - a[j]] > i && pos[a[i] - a[j]] < R[i]) ans++;
    } else {
      for (long long j = i + 1; j <= R[i] - 1; j++)
        if (pos[a[i] - a[j]] < i && pos[a[i] - a[j]] > L[i]) ans++;
    }
  }
  printf("%lld\n", ans);
  return 0;
}
