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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,avx,avx2")
using namespace std;
mt19937 rnd(time(0));
const int INF = 1000 * 1000 * 1000 + 7;
const long long LINF = 1LL * INF * INF;
const int MAX = 1e5 + 7;
const double PI = acos(-1.);
const double EPS = 1e-8;
const int MOD = 1000 * 1000 * 1000 + 7;
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  vector<int> p(n + 1), pos(n + 1);
  for (int i = 1; i < n + 1; ++i) {
    cin >> p[i];
    pos[p[i]] = i;
  }
  stack<pair<int, int> > st;
  vector<int> sprava(n + 1), sleva(n + 1);
  st.push({INF, 0});
  for (int i = 1; i < n + 1; ++i) {
    while (st.top().first < p[i]) {
      st.pop();
    }
    sleva[i] = st.top().second;
    st.push({p[i], i});
  }
  while (!st.empty()) {
    st.pop();
  }
  st.push({INF, n + 1});
  for (int i = n + 1 - 1; i >= 1; --i) {
    while (st.top().first < p[i]) {
      st.pop();
    }
    sprava[i] = st.top().second;
    st.push({p[i], i});
  }
  long long ans = 0;
  for (int i = 2; i < n; ++i) {
    if (sprava[i] - sleva[i] > 1) {
      if (i - sleva[i] < sprava[i] - i) {
        for (int j = i - 1; j >= sleva[i] + 1; --j) {
          if (pos[p[i] - p[j]] < sprava[i] && pos[p[i] - p[j]] > i) {
            ++ans;
          }
        }
      } else {
        for (int j = sprava[i] - 1; j >= i + 1; --j) {
          if (pos[p[i] - p[j]] > sleva[i] && pos[p[i] - p[j]] < i) {
            ++ans;
          }
        }
      }
    }
  }
  cout << ans;
  return 0;
}
