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 maxn = 200001;
int a[maxn];
int n;
void enter() {
  cin >> n;
  for (int i = (1), _b = (n + 1); i < _b; ++i) cin >> a[i];
}
int l[maxn], r[maxn];
int pos[maxn];
void solve() {
  deque<int> kew;
  for (int i = (1), _b = (n + 1); i < _b; ++i) {
    while (!kew.empty() && a[kew.back()] < a[i]) kew.pop_back();
    if (kew.empty())
      l[i] = 1;
    else
      l[i] = kew.back() + 1;
    kew.push_back(i);
  }
  kew.clear();
  for (int i = n; i >= 1; --i) {
    while (!kew.empty() && a[kew.back()] < a[i]) kew.pop_back();
    if (kew.empty())
      r[i] = n;
    else
      r[i] = kew.back() - 1;
    kew.push_back(i);
  }
  long long ans = 0;
  for (int i = (1), _b = (n + 1); i < _b; ++i) pos[a[i]] = i;
  for (int i = (2), _b = (n); i < _b; ++i) {
    if (i - l[i] < r[i] - i) {
      for (int j = l[i]; j < i; ++j)
        if (pos[a[i] - a[j]] > i && pos[a[i] - a[j]] <= r[i]) ++ans;
    } else {
      for (int j = i + 1; j <= r[i]; ++j)
        if (pos[a[i] - a[j]] >= l[i] && pos[a[i] - a[j]] < i) ++ans;
    }
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie();
  cout.tie();
  enter();
  solve();
}
