#include <bits/stdc++.h>
using namespace std;
int x, minn, n, a[1000001], k;
int main() {
  scanf("%d", &n);
  minn = 1000000001;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    minn = min(minn, a[i]);
  }
  k = minn;
  minn = 1000000001;
  for (int i = 0; i < n; i++) {
    if (a[i] != k) continue;
    if (x != 0) minn = min(minn, i + 1 - x);
    x = i + 1;
  }
  printf("%d\n", minn);
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 x, minn, n, a[1000001], k;
int main() {
  scanf("%d", &n);
  minn = 1000000001;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    minn = min(minn, a[i]);
  }
  k = minn;
  minn = 1000000001;
  for (int i = 0; i < n; i++) {
    if (a[i] != k) continue;
    if (x != 0) minn = min(minn, i + 1 - x);
    x = i + 1;
  }
  printf("%d\n", minn);
}

```