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;
template <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
const long long maxn = 1000001, mod = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int ar[n];
  map<int, int> mapp;
  for (long long i = 0; i < (long long)(n); ++i) {
    cin >> ar[i];
    mapp[ar[i]] = i;
  }
  set<int> s;
  s.insert(-1);
  s.insert(n);
  int ans = 0;
  for (int i = n; i >= 1; i--) {
    int idx = mapp[i];
    auto it = s.lower_bound(idx);
    auto iit = s.upper_bound(idx);
    if (it != s.begin()) it--;
    int a = *it + 1;
    int b = *iit - 1;
    if (idx - a < b - idx) {
      for (long long j = (long long)(a); j <= (long long)(idx - 1); ++j) {
        int acha = mapp[i - ar[j]];
        if (acha > idx && acha <= b) {
          ans++;
        }
      }
    } else {
      for (long long j = (long long)(idx + 1); j <= (long long)(b); ++j) {
        int acha = mapp[i - ar[j]];
        if (acha < idx && acha >= a) {
          ans++;
        }
      }
    }
    s.insert(mapp[i]);
  }
  cout << ans;
  return 0;
}
