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 = 3e5;
int n, ans, seg[N * 4];
vector<int> v, pos;
void build(int p, int l, int r) {
  if (l == r) {
    seg[p] = v[l];
    return;
  }
  int mid = (l + r) / 2;
  build(p * 2, l, mid), build(p * 2 + 1, mid + 1, r);
  seg[p] = max(seg[p * 2], seg[p * 2 + 1]);
}
int get(int p, int l, int r, int s, int e) {
  if (l > e || r < s) return 0;
  if (l >= s && r <= e) return seg[p];
  int mid = (l + r) / 2;
  return max(get(p * 2, l, mid, s, e), get(p * 2 + 1, mid + 1, r, s, e));
}
void calc(int l, int r) {
  if (l >= r) return;
  int mx = get(1, 0, n - 1, l, r);
  int posMax = pos[mx];
  int grateL = r - posMax;
  int grateR = posMax - l;
  if (grateL < grateR) {
    for (int i = posMax + 1; i <= r; i++)
      if (v[i] < mx && pos[mx - v[i]] < posMax && pos[mx - v[i]] >= l) ans++;
  } else {
    for (int i = l; i <= posMax - 1; i++)
      if (v[i] < mx && pos[mx - v[i]] > posMax && pos[mx - v[i]] <= r) ans++;
  }
  calc(l, posMax - 1), calc(posMax + 1, r);
}
int main() {
  scanf("%d", &n);
  v.resize(n), pos.resize(n + 10);
  for (int i = 0; i < n; i++) scanf("%d", &v[i]), pos[v[i]] = i;
  build(1, 0, n - 1);
  calc(0, n - 1);
  printf("%d", ans);
  return 0;
}
