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;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> p(n);
  for (auto &t : p) {
    cin >> t;
  }
  vector<int> pos(n + 1);
  for (int i = 0; i < n; i++) {
    pos[p[i]] = i;
  }
  vector<int> pr(n);
  {
    vector<int> st;
    for (int i = n - 1; i >= 0; i--) {
      while (!st.empty() && p[i] > p[st.back()]) {
        st.pop_back();
      }
      if (st.empty()) {
        pr[i] = n;
      } else {
        pr[i] = st.back();
      }
      st.push_back(i);
    }
  }
  vector<int> pl(n);
  {
    vector<int> st;
    for (int i = 0; i < n; i++) {
      while (!st.empty() && p[i] > p[st.back()]) {
        st.pop_back();
      }
      if (st.empty()) {
        pl[i] = -1;
      } else {
        pl[i] = st.back();
      }
      st.push_back(i);
    }
  }
  if (false) cerr << "init ok" << endl;
  int ans = 0;
  for (int i = 0; i < n; i++) {
    int l1 = (i - pl[i]);
    int l2 = (pr[i] - i);
    assert(i < pr[i]);
    assert(pl[i] < i);
    if (l1 < l2) {
      if (false) cerr << "from " << pl[i] + 1 << " to " << i - 1 << endl;
      for (int j = pl[i] + 1; j < i; j++) {
        int need = p[i] - p[j];
        if (need > 0 && (i < pos[need] && pos[need] < pr[i])) {
          ans++;
        }
      }
    } else {
      if (false) cerr << "from " << i + 1 << "  to " << pr[i] - 1 << endl;
      for (int j = i + 1; j < pr[i]; j++) {
        int need = p[i] - p[j];
        if (need > 0 && (pl[i] < pos[need] && pos[need] < i)) {
          ans++;
        }
      }
    }
  }
  cout << ans << endl;
}
