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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  long long mas[n + 1], pos[n + 1];
  pair<long long, long long> bord[n + 1];
  stack<pair<long long, long long> > st;
  for (long long i = 1; i <= n; i++) {
    bord[i] = {1, n};
    cin >> mas[i];
    pos[mas[i]] = i;
    while (!st.empty() && st.top().first < mas[i]) st.pop();
    if (!st.empty()) bord[i].first = st.top().second + 1;
    st.push({mas[i], i});
  }
  while (!st.empty()) st.pop();
  for (long long i = n; i > 0; i--) {
    while (!st.empty() && st.top().first < mas[i]) st.pop();
    if (!st.empty()) bord[i].second = st.top().second - 1;
    st.push({mas[i], i});
  }
  long long res = 0;
  for (long long i = 1; i <= n; i++) {
    long long l = bord[i].first, r = bord[i].second;
    long long ml = l, mr = i - 1, sl = i + 1, sr = r;
    if (i - l > r - i) ml = i + 1, mr = r, sl = l, sr = i - 1;
    for (long long j = ml; j <= mr; j++)
      res += (sl <= pos[mas[i] - mas[j]] && pos[mas[i] - mas[j]] <= sr);
  }
  cout << res << "\n";
  return 0;
}
