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>
const int MAXN = 1e6;
const int INF = 0x3f3f3f3f;
using namespace std;
inline void read(int &x) {
  x = 0;
  char c = getchar();
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  }
}
int n, a[MAXN + 5], h[MAXN + 5];
void init() {
  read(n);
  for (int i = 1; i <= n; ++i) read(a[i]);
}
stack<int> sta;
pair<int, int> seg[MAXN + 5];
int main() {
  init();
  a[0] = INF, a[n + 1] = INF, sta.push(0);
  for (int i = 1; i <= n + 1; ++i) {
    while (!sta.empty() && a[sta.top()] < a[i]) {
      seg[sta.top()].second = i - 1, sta.pop();
    }
    seg[i].first = sta.top() + 1, sta.push(i);
    if (i <= n) h[a[i]] = i;
  }
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    int l1 = seg[i].first, r1 = i - 1, l2 = i + 1, r2 = seg[i].second;
    if (r1 - l1 > r2 - l2) {
      swap(l1, l2), swap(r1, r2);
    }
    for (int j = l1; j <= r1; ++j) {
      int pos = h[a[i] - a[j]];
      if (l2 <= pos && pos <= r2) ++ans;
    }
  }
  printf("%I64d\n", ans);
  return 0;
}
