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 long long N = 5e5 + 3;
const long long M = 1e6 + 4;
const long long INF = 1e9 + 7;
const int SM = 1e3 + 4;
int n, a[N], pos[N], l[N], r[N], ans = 0;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    pos[a[i]] = i;
  }
  stack<int> st;
  for (int i = 1; i <= n; i++) {
    l[i] = 1;
    while (!st.empty() && a[st.top()] <= a[i]) {
      st.pop();
    }
    if (!st.empty()) {
      l[i] = st.top() + 1;
    }
    st.push(i);
  }
  while (!st.empty()) {
    st.pop();
  }
  for (int i = n; i >= 1; i--) {
    r[i] = n;
    while (!st.empty() && a[st.top()] <= a[i]) {
      st.pop();
    }
    if (!st.empty()) {
      r[i] = st.top() - 1;
    }
    st.push(i);
  }
  for (int i = 1; i <= n; i++) {
    int len1 = i - l[i] + 1;
    int len2 = r[i] - i + 1;
    if (len1 <= len2) {
      for (int L = l[i]; L < i; L++) {
        int x = a[i] - a[L];
        int R = pos[x];
        if (i <= R && R <= r[i]) {
          ans++;
        }
      }
    } else {
      for (int R = i + 1; R <= r[i]; R++) {
        int x = a[i] - a[R];
        int L = pos[x];
        if (l[i] <= L && L <= i) {
          ans++;
        }
      }
    }
  }
  cout << ans;
  return 0;
}
