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 n, p[200005], er[200005], Mm[200005][2];
stack<int> s;
int DC(int l, int r) {
  if (l == r) return 0;
  int m = (l + r) / 2, tp = 0, mxi = 0;
  for (int i = m + 1; i <= r; ++i) {
    if (p[i] > p[mxi]) mxi = i;
    int si = er[p[mxi] - p[i]];
    if (Mm[mxi][0] < si && si >= l && si <= m) ++tp;
  }
  mxi = 0;
  for (int i = m; i >= l; --i) {
    if (p[i] > p[mxi]) mxi = i;
    int si = er[p[mxi] - p[i]];
    if (Mm[mxi][1] > si && si > m && si <= r) ++tp;
  }
  return tp + DC(l, m) + DC(m + 1, r);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> p[i];
    er[p[i]] = i;
  }
  for (int i = 1; i <= n; ++i) {
    while (s.size() && p[s.top()] < p[i]) s.pop();
    if (s.size()) Mm[i][0] = s.top();
    s.push(i);
  }
  while (s.size()) s.pop();
  for (int i = n; i >= 1; --i) {
    while (s.size() && p[s.top()] < p[i]) s.pop();
    if (s.size())
      Mm[i][1] = s.top();
    else
      Mm[i][1] = 1e9;
    s.push(i);
  }
  cout << DC(1, n) << endl;
}
