#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, j = 0, k = 0;
  cin >> n;
  int a[n], b[n], c[n], ans[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(b, b + n);
  int x = b[0];
  for (int i = 0; i < n; i++) {
    if (a[i] == x) {
      c[j] = i;
      if (j > 0) {
        ans[k] = c[j] - c[j - 1];
        k++;
      }
      j++;
    }
  }
  sort(ans, ans + k);
  cout << ans[0] << endl;
}

### Prompt

Create a solution in CPP for the following problem:

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() {
  int n, j = 0, k = 0;
  cin >> n;
  int a[n], b[n], c[n], ans[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(b, b + n);
  int x = b[0];
  for (int i = 0; i < n; i++) {
    if (a[i] == x) {
      c[j] = i;
      if (j > 0) {
        ans[k] = c[j] - c[j - 1];
        k++;
      }
      j++;
    }
  }
  sort(ans, ans + k);
  cout << ans[0] << endl;
}

```