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 inf = 0x3f3f3f3f;
const int maxn = 222229;
int vis[maxn];
int a[maxn];
int dp[maxn][26], n;
void RMQ() {
  for (int i = 1; i <= n; i++) dp[i][0] = a[i];
  for (int j = 1; (1 << j) <= n; j++)
    for (int i = 1; i + (1 << j) - 1 <= n; i++)
      dp[i][j] = max(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
}
int finds(int l, int r) {
  int h = log2(r - l + 1);
  return max(dp[l][h], dp[r - (1 << h) + 1][h]);
}
int dfs(int l, int r) {
  if (l >= r) return 0;
  int ans = 0;
  int ret = finds(l, r);
  int p = vis[ret];
  ans += dfs(l, p - 1);
  ans += dfs(p + 1, r);
  if (p - l < r - p) {
    for (int i = l; i <= p - 1; i++)
      if (vis[ret - a[i]] >= p + 1 && vis[ret - a[i]] <= r) ans++;
  } else {
    for (int i = p + 1; i <= r; i++)
      if (vis[ret - a[i]] >= l && vis[ret - a[i]] <= p - 1) ans++;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    vis[a[i]] = i;
  }
  RMQ();
  cout << dfs(1, n) << endl;
  return 0;
}
