#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  int min = 1000000001;
  int num;
  int minIndex;
  vector<int> ar;
  for (int i = 0; i < n; i++) {
    cin >> num;
    ar.push_back(num);
    if (num < min) {
      min = num;
    }
  }
  int co = 0;
  long long ans = 1000000001;
  for (int i = 0; i < n; i++) {
    if (min == ar[i]) {
      if (co > 0) {
        if (ans > i - minIndex) {
          ans = i - minIndex;
        }
      }
      minIndex = i;
      co++;
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please create a solution in cpp 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() {
  int n;
  cin >> n;
  int min = 1000000001;
  int num;
  int minIndex;
  vector<int> ar;
  for (int i = 0; i < n; i++) {
    cin >> num;
    ar.push_back(num);
    if (num < min) {
      min = num;
    }
  }
  int co = 0;
  long long ans = 1000000001;
  for (int i = 0; i < n; i++) {
    if (min == ar[i]) {
      if (co > 0) {
        if (ans > i - minIndex) {
          ans = i - minIndex;
        }
      }
      minIndex = i;
      co++;
    }
  }
  cout << ans << endl;
  return 0;
}

```