#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  int a[100000];
  int mins[100000];
  int mini;
  cin >> n;
  cin >> a[0];
  mini = a[0];
  for (int i = 1; i < n; i++) {
    cin >> a[i];
    if (a[i] < mini) mini = a[i];
  }
  int j = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] == mini) {
      mins[j] = i;
      j++;
    }
  }
  int distance = n;
  for (int i = 1; i < j; i++) {
    if (mins[i] - mins[i - 1] < distance) distance = mins[i] - mins[i - 1];
  }
  cout << distance;
}

### 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() {
  int n;
  int a[100000];
  int mins[100000];
  int mini;
  cin >> n;
  cin >> a[0];
  mini = a[0];
  for (int i = 1; i < n; i++) {
    cin >> a[i];
    if (a[i] < mini) mini = a[i];
  }
  int j = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] == mini) {
      mins[j] = i;
      j++;
    }
  }
  int distance = n;
  for (int i = 1; i < j; i++) {
    if (mins[i] - mins[i - 1] < distance) distance = mins[i] - mins[i - 1];
  }
  cout << distance;
}

```