#include <bits/stdc++.h>
int n;
int a[200000];
int main() {
  scanf("%d", &n);
  int min = 2147483647;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (min > a[i]) min = a[i];
  }
  int now = -1000000011, ans = 2147483647;
  for (int i = 0; i < n; i++)
    if (a[i] == min) {
      if (i - now < ans) ans = i - now;
      now = i;
    }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please formulate 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>
int n;
int a[200000];
int main() {
  scanf("%d", &n);
  int min = 2147483647;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (min > a[i]) min = a[i];
  }
  int now = -1000000011, ans = 2147483647;
  for (int i = 0; i < n; i++)
    if (a[i] == min) {
      if (i - now < ans) ans = i - now;
      now = i;
    }
  printf("%d\n", ans);
  return 0;
}

```