#include <bits/stdc++.h>
using namespace std;
int n, i, ans1 = 1000000001, ans11, ans = 1000000001;
int a[100001];
int main() {
  cin >> n;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= n; i++) {
    if (ans1 > a[i]) {
      ans1 = a[i];
      ans11 = i;
    }
  }
  for (i = 1; i <= n; i++) {
    if (a[i] == a[ans11] && i != ans11) {
      ans = min(ans, abs(ans11 - i));
      ans11 = i;
    }
  }
  cout << ans << 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 n, i, ans1 = 1000000001, ans11, ans = 1000000001;
int a[100001];
int main() {
  cin >> n;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= n; i++) {
    if (ans1 > a[i]) {
      ans1 = a[i];
      ans11 = i;
    }
  }
  for (i = 1; i <= n; i++) {
    if (a[i] == a[ans11] && i != ans11) {
      ans = min(ans, abs(ans11 - i));
      ans11 = i;
    }
  }
  cout << ans << endl;
}

```