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 MAXN = 2e5 + 5;
const int INF = 1e9;
int L[MAXN], R[MAXN], p[MAXN], pos[MAXN];
stack<int> st;
long long calc(int l, int r, int pl, int pr, int mx) {
  long long ret = 0;
  for (int i = l; i <= r; i++) {
    if (pos[mx - p[i]] >= pl && pos[mx - p[i]] <= pr) ret++;
  }
  return ret;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
    pos[p[i]] = i;
  }
  p[0] = p[n + 1] = INF;
  st.push(0);
  for (int i = 1; i <= n; i++) {
    while (p[i] > p[st.top()]) st.pop();
    L[i] = st.top();
    st.push(i);
  }
  while (!st.empty()) st.pop();
  st.push(n + 1);
  for (int i = n; i >= 1; i--) {
    while (p[i] > p[st.top()]) st.pop();
    R[i] = st.top();
    st.push(i);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int lenL = i - L[i];
    int lenR = R[i] - i;
    if (lenL > lenR)
      ans += calc(i + 1, R[i] - 1, L[i] + 1, i - 1, p[i]);
    else
      ans += calc(L[i] + 1, i - 1, i + 1, R[i] - 1, p[i]);
  }
  printf("%lld\n", ans);
  return 0;
}
