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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> elements(n);
  vector<int> element_position_map(n + 1);
  for (size_t i = 0; i < n; i++) {
    cin >> elements[i];
    element_position_map[elements[i]] = i;
  }
  deque<pair<int, int>> current_elements;
  vector<long long> left_count(n), right_count(n);
  for (size_t i = 0; i < n; i++) {
    int new_element = elements[i];
    int new_count = 0;
    while (!current_elements.empty() &&
           elements[current_elements.back().first] < new_element) {
      new_count++;
      new_count += current_elements.back().second;
      right_count[current_elements.back().first] =
          i - current_elements.back().first - 1;
      current_elements.pop_back();
    }
    left_count[i] = new_count;
    current_elements.push_back({i, new_count});
  }
  while (!current_elements.empty()) {
    right_count[current_elements.back().first] =
        n - current_elements.back().first - 1;
    current_elements.pop_back();
  }
  long long solution = 0;
  for (size_t i = 0; i < n; i++) {
    if (left_count[i] == 0 || right_count[i] == 0) continue;
    pair<int, int> segment_to_iterate = {i - left_count[i], i - 1},
                   other_segment = {i + 1, i + right_count[i]};
    if (right_count[i] < left_count[i]) swap(segment_to_iterate, other_segment);
    int max_element_value = elements[i];
    for (size_t j = segment_to_iterate.first; j <= segment_to_iterate.second;
         j++) {
      int other_element = max_element_value - elements[j];
      int other_element_position = element_position_map[other_element];
      if (other_element_position >= other_segment.first &&
          other_element_position <= other_segment.second)
        solution++;
    }
  }
  cout << solution;
  return 0;
}
