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;
mt19937 bruh(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rofl(chrono::steady_clock::now().time_since_epoch().count());
const int N = 2e5 + 2;
const int M = 1e5;
const int mod = 0;
const int inf = 2e9 + 3;
const long long INF = 1e18;
const int dx[4] = {1, -1, 0, 0};
const int dy[4] = {0, 0, -1, 1};
void files() {
  freopen(".in", "r", stdin);
  freopen(".out", "w", stdout);
}
int n, a[N], p[N], l[N], r[N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    p[a[i]] = i;
  }
  stack<pair<int, int> > c;
  c.push({inf, 0});
  for (int i = 1; i <= n; ++i) {
    while (c.top().first < a[i]) c.pop();
    l[a[i]] = c.top().second;
    c.push({a[i], i});
  }
  while ((int)c.size()) c.pop();
  c.push({inf, n + 1});
  for (int i = n; i >= 1; --i) {
    while (c.top().first < a[i]) c.pop();
    r[a[i]] = c.top().second;
    c.push({a[i], i});
  }
  while ((int)c.size()) c.pop();
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    if (p[i] - l[i] < r[i] - p[i])
      for (int j = l[i] + 1; j < p[i]; ++j)
        ans += (p[i] < p[i - a[j]] && p[i - a[j]] < r[i]);
    else
      for (int j = p[i] + 1; j < r[i]; ++j)
        ans += (l[i] < p[i - a[j]] && p[i - a[j]] < p[i]);
  }
  printf("%lld", ans);
}
