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;
long long ans, n;
long long l[200002], r[200002], a[200002], pos[200002];
void calc(long long l, long long r, long long l1, long long r1, long long p) {
  for (int i = l; i <= r; i++) {
    long long rem = a[p] - a[i];
    if (pos[rem] >= l1 && pos[rem] <= r1) ans++;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(), cout.tie();
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    pos[a[i]] = i;
  }
  stack<pair<long long, long long>> st;
  st.push({n + 1, -1});
  for (int i = 0; i < n; i++) {
    while (st.top().first < a[i]) st.pop();
    l[i] = st.top().second + 1;
    st.push({a[i], i});
  }
  while (!st.empty()) st.pop();
  st.push({n + 1, n});
  for (int i = n - 1; i >= 0; i--) {
    while (st.top().first < a[i]) st.pop();
    r[i] = st.top().second - 1;
    st.push({a[i], i});
  }
  for (int i = 0; i < n; i++) {
    if ((i - l[i]) < 1 || (r[i] - i) < 1) continue;
    if ((i - l[i]) <= (r[i] - i))
      calc(l[i], i - 1, i + 1, r[i], i);
    else
      calc(i + 1, r[i], l[i], i - 1, i);
  }
  cout << ans << endl;
  return 0;
}
