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>
const int N = 200000;
int n, a[N], pos[N], l[N], r[N];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
    --a[i];
    pos[a[i]] = i;
  }
  std::vector<int> st;
  for (int i = 0; i < n; ++i) {
    while (!st.empty() && a[st.back()] < a[i]) st.pop_back();
    if (st.empty())
      l[i] = 0;
    else
      l[i] = st.back() + 1;
    st.push_back(i);
  }
  st.clear();
  for (int i = n - 1; i >= 0; --i) {
    while (!st.empty() && a[st.back()] < a[i]) st.pop_back();
    if (st.empty())
      r[i] = n - 1;
    else
      r[i] = st.back() - 1;
    st.push_back(i);
  }
  int ans = 0;
  for (int i = 0; i < n; ++i) {
    if (i - l[i] < r[i] - i) {
      for (int j = l[i]; j < i; ++j)
        if (pos[a[i] - a[j] - 1] > i && pos[a[i] - a[j] - 1] <= r[i]) ++ans;
    } else
      for (int j = i + 1; j <= r[i]; ++j)
        if (pos[a[i] - a[j] - 1] < i && pos[a[i] - a[j] - 1] >= l[i]) ++ans;
  }
  printf("%d\n", ans);
  return 0;
}
