#include <bits/stdc++.h>
using namespace std;
int main() {
  int a, b, c, n, arr[100005];
  int mn = INT_MAX;
  int cnt = 0;
  vector<int> vec;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    mn = min(mn, arr[i]);
  }
  for (int i = 0; i < n; i++) {
    if (arr[i] == mn) vec.push_back(i);
  }
  int mnPos = INT_MAX;
  for (int i = 1; i < vec.size(); i++) {
    mnPos = min(mnPos, vec[i] - vec[i - 1]);
  }
  cout << mnPos << endl;
}

### 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;
int main() {
  int a, b, c, n, arr[100005];
  int mn = INT_MAX;
  int cnt = 0;
  vector<int> vec;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    mn = min(mn, arr[i]);
  }
  for (int i = 0; i < n; i++) {
    if (arr[i] == mn) vec.push_back(i);
  }
  int mnPos = INT_MAX;
  for (int i = 1; i < vec.size(); i++) {
    mnPos = min(mnPos, vec[i] - vec[i - 1]);
  }
  cout << mnPos << endl;
}

```