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;
const int MAXN = 200005;
int p[MAXN], loc[MAXN];
struct DSU {
  int fa[MAXN];
  set<int> val[MAXN];
  void init(int n) {
    for (int i = 1; i <= n; i++) {
      fa[i] = i;
      val[i].insert(p[i]);
    }
  }
  int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
  void merge(int x, int y) {
    x = find(x), y = find(y);
    if (x == y) return;
    fa[x] = y;
    if (val[x].size() > val[y].size()) val[x].swap(val[y]);
    for (auto &t : val[x]) val[y].insert(t);
    val[x].clear();
  }
  int query(int x, int y, int v) {
    x = find(x), y = find(y);
    if (val[x].size() > val[y].size()) swap(x, y);
    int res = 0;
    for (auto &t : val[x]) res += val[y].count(v - t);
    return res;
  }
} dsu;
bool vis[MAXN];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]), loc[p[i]] = i;
  dsu.init(n);
  int res = 0;
  for (int i = 1; i <= n; i++) {
    if (vis[loc[i] - 1] && vis[loc[i] + 1])
      res += dsu.query(loc[i] - 1, loc[i] + 1, i);
    if (vis[loc[i] - 1]) dsu.merge(loc[i] - 1, loc[i]);
    if (vis[loc[i] + 1]) dsu.merge(loc[i], loc[i] + 1);
    vis[loc[i]] = 1;
  }
  printf("%d\n", res);
  return 0;
}
