#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, i, j;
  cin >> n;
  long long int a[n], te = 100000000000, dis, count = 0, p = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (te > a[i]) te = a[i];
  }
  dis = n + 1;
  for (i = 0; i < n; i++) {
    if (te == a[i]) {
      p++;
      if (dis > count && p > 1) {
        dis = count;
      }
      count = 0;
    } else
      count++;
  }
  cout << dis + 1 << endl;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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, i, j;
  cin >> n;
  long long int a[n], te = 100000000000, dis, count = 0, p = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (te > a[i]) te = a[i];
  }
  dis = n + 1;
  for (i = 0; i < n; i++) {
    if (te == a[i]) {
      p++;
      if (dis > count && p > 1) {
        dis = count;
      }
      count = 0;
    } else
      count++;
  }
  cout << dis + 1 << endl;
  return 0;
}

```