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;
vector<int> dsu, whereset;
vector<set<int>> se;
void build(int n) {
  dsu.resize(n);
  whereset.resize(n);
  se.resize(n);
  for (int i = 0; i < n; i++) {
    dsu[i] = i;
    se[i].insert(i);
    whereset[i] = i;
  }
}
int Find(int a) { return (dsu[a] == a ? a : dsu[a] = Find(dsu[a])); }
long long ans = 0;
void Union(int a, int b, int val) {
  a = Find(a);
  b = Find(b);
  if (rand() % 2) swap(a, b);
  dsu[b] = a;
  dsu[val] = a;
  if (se[whereset[a]].size() < se[whereset[b]].size()) {
    for (int i : se[whereset[a]]) {
      if (se[whereset[b]].find(val - i - 1) != se[whereset[b]].end()) ans++;
    }
    for (int i : se[whereset[a]]) se[whereset[b]].insert(i);
    se[whereset[b]].insert(val);
    whereset[a] = whereset[b];
  } else {
    for (int i : se[whereset[b]]) {
      if (se[whereset[a]].find(val - i - 1) != se[whereset[a]].end()) ans++;
    }
    for (int i : se[whereset[b]]) se[whereset[a]].insert(i);
    se[whereset[a]].insert(val);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> p(n), pos(n);
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
    pos[p[i]] = i;
  }
  build(n);
  for (int i = 0; i < n; i++) {
    if (pos[i] > 0 && pos[i] < n - 1 && p[pos[i] + 1] < i &&
        p[pos[i] - 1] < i) {
      Union(p[pos[i] - 1], p[pos[i] + 1], i);
    } else if (pos[i] < n - 1 && p[pos[i] + 1] < i) {
      dsu[i] = Find(dsu[p[pos[i] + 1]]);
      se[whereset[dsu[i]]].insert(i);
    } else if (pos[i] > 0 && p[pos[i] - 1] < i) {
      dsu[i] = Find(dsu[p[pos[i] - 1]]);
      se[whereset[dsu[i]]].insert(i);
    }
  }
  cout << ans;
}
