#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  vector<int> a(n);
  bool flag = false;
  int prev, mn;
  int ans = n + 1;
  mn = (1000000007);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] < mn) {
      mn = a[i];
      prev = i;
      ans = n + 1;
      flag = true;
    } else if (a[i] == mn) {
      ans = min(ans, i - prev);
      prev = i;
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

You are given an array of n integer numbers a0, a1, ..., an - 1. Find the distance between two closest (nearest) minimums in it. It is guaranteed that in the array a minimum occurs at least two times.

Input

The first line contains positive integer n (2 ≤ n ≤ 105) — size of the given array. The second line contains n integers a0, a1, ..., an - 1 (1 ≤ ai ≤ 109) — elements of the array. It is guaranteed that in the array a minimum occurs at least two times.

Output

Print the only number — distance between two nearest minimums in the array.

Examples

Input

2
3 3


Output

1


Input

3
5 6 5


Output

2


Input

9
2 1 3 5 4 1 2 3 1


Output

3

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  vector<int> a(n);
  bool flag = false;
  int prev, mn;
  int ans = n + 1;
  mn = (1000000007);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] < mn) {
      mn = a[i];
      prev = i;
      ans = n + 1;
      flag = true;
    } else if (a[i] == mn) {
      ans = min(ans, i - prev);
      prev = i;
    }
  }
  cout << ans << endl;
  return 0;
}

```