#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n;
  cin >> n;
  vector<long long int> arr(n);
  long long int mi = INT_MAX;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    mi = min(mi, arr[i]);
  }
  long long int ans = INT_MAX, idx = -1;
  for (int i = 0; i < n; i++) {
    if (arr[i] == mi) {
      if (idx == -1)
        idx = i;
      else {
        ans = min(ans, i - idx);
        idx = i;
      }
    }
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  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;
void solve() {
  int n;
  cin >> n;
  vector<long long int> arr(n);
  long long int mi = INT_MAX;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    mi = min(mi, arr[i]);
  }
  long long int ans = INT_MAX, idx = -1;
  for (int i = 0; i < n; i++) {
    if (arr[i] == mi) {
      if (idx == -1)
        idx = i;
      else {
        ans = min(ans, i - idx);
        idx = i;
      }
    }
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}

```