#include <bits/stdc++.h>
using namespace std;
int maxn = 1e9 + 10;
int main() {
  int n;
  cin >> n;
  int minn = maxn;
  int ans = maxn;
  int pre = 0;
  for (int i = 0; i < n; i++) {
    int num;
    cin >> num;
    pre++;
    if (num < minn) {
      minn = num;
      ans = maxn;
      pre = 0;
      continue;
    }
    if (num == minn) {
      if (pre < ans) ans = pre;
      pre = 0;
      continue;
    }
  }
  cout << ans << endl;
}

### Prompt

Please provide a Cpp coded solution 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 maxn = 1e9 + 10;
int main() {
  int n;
  cin >> n;
  int minn = maxn;
  int ans = maxn;
  int pre = 0;
  for (int i = 0; i < n; i++) {
    int num;
    cin >> num;
    pre++;
    if (num < minn) {
      minn = num;
      ans = maxn;
      pre = 0;
      continue;
    }
    if (num == minn) {
      if (pre < ans) ans = pre;
      pre = 0;
      continue;
    }
  }
  cout << ans << endl;
}

```