#include <bits/stdc++.h>
using namespace std;
int n, in, c;
int a[100010], mn = 1e9, ans = 1e9;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (a[i] < mn) {
      mn = a[i];
      in = i;
    }
  }
  for (in += 1; in < n; in++) {
    c++;
    if (a[in] == mn) {
      ans = min(ans, c);
      c = 0;
    }
  }
  printf("%d", ans);
  return 0;
}

### Prompt

Your task is to create a cpp solution to 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 n, in, c;
int a[100010], mn = 1e9, ans = 1e9;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (a[i] < mn) {
      mn = a[i];
      in = i;
    }
  }
  for (in += 1; in < n; in++) {
    c++;
    if (a[in] == mn) {
      ans = min(ans, c);
      c = 0;
    }
  }
  printf("%d", ans);
  return 0;
}

```