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 N = 4e5 + 10, OO = 1e9, Mod = 1e9 + 7;
int n, a[N];
int idx[N];
int lefts[N], rights[N];
stack<pair<int, int>> st;
void rem() {
  while (st.size()) st.pop();
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    idx[a[i]] = i;
  }
  for (int i = 1; i <= n; ++i) {
    while (!st.empty() && a[i] >= st.top().first) {
      st.pop();
    }
    if (st.empty())
      lefts[i] = 0;
    else
      lefts[i] = st.top().second;
    st.push({a[i], i});
  }
  rem();
  for (int i = n; i >= 1; --i) {
    while (!st.empty() && a[i] >= st.top().first) {
      st.pop();
    }
    if (st.empty()) {
      rights[i] = n + 1;
    } else {
      rights[i] = st.top().second;
    }
    st.push({a[i], i});
  }
  rem();
  int ans = 0;
  for (int i = 2; i < n; ++i) {
    int rangeleft = i - lefts[i];
    int rangeright = rights[i] - i;
    if (rangeleft >= rangeright) {
      for (int j = i + 1; j < rights[i]; ++j) {
        int need = a[i] - a[j];
        if (idx[need] > lefts[i] && idx[need] < i) ++ans;
      }
    } else {
      for (int j = i - 1; j > lefts[i]; --j) {
        int need = a[i] - a[j];
        if (idx[need] > i && idx[need] < rights[i]) ++ans;
      }
    }
  }
  cout << ans << endl;
}
