#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n;
  cin >> n;
  long long a[n];
  long long mn = 1e15;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    mn = min(a[i], mn);
  }
  long long l = -1;
  long long ans = 1e7;
  for (long long i = 0; i < n; i++) {
    if (a[i] == mn) {
      if (l != -1) {
        ans = min(ans, i - l);
      }
      l = i;
    }
  }
  cout << ans;
}

### Prompt

Your challenge is to write 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 main() {
  long long n;
  cin >> n;
  long long a[n];
  long long mn = 1e15;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    mn = min(a[i], mn);
  }
  long long l = -1;
  long long ans = 1e7;
  for (long long i = 0; i < n; i++) {
    if (a[i] == mn) {
      if (l != -1) {
        ans = min(ans, i - l);
      }
      l = i;
    }
  }
  cout << ans;
}

```