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 = 0;
long long pos[1000000];
long long a[1000000];
void calc(long long l, long long r, long long l1, long long r1, long long md) {
  for (long long i = l; i <= r; i++)
    if (pos[a[md] - a[i]] >= l1 && pos[a[md] - a[i]] <= r1) ans++;
}
int main() {
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i];
  stack<pair<long long, long long> > st;
  long long left[n], right[n];
  for (long long i = 0; i < n; i++) {
    while (st.size() > 0 && st.top().first < a[i]) st.pop();
    if (st.size() == 0)
      left[i] = -1;
    else
      left[i] = st.top().second;
    st.push({a[i], i});
  }
  while (st.size() > 0) st.pop();
  for (long long i = n - 1; i >= 0; i--) {
    while (st.size() > 0 && st.top().first < a[i]) st.pop();
    if (st.size() == 0)
      right[i] = n;
    else
      right[i] = st.top().second;
    st.push({a[i], i});
  }
  for (long long i = 0; i < n; i++) left[i]++, right[i]--;
  for (long long i = 0; i < n; i++) pos[a[i]] = i;
  for (long long i = 0; i < n; i++)
    if (i - left[i] < right[i] - i)
      calc(left[i], i - 1, i + 1, right[i], i);
    else
      calc(i + 1, right[i], left[i], i - 1, i);
  cout << ans;
}
