#include <bits/stdc++.h>
using namespace std;
map<int, int> M;
int n;
int main() {
  cin >> n;
  int res = 1e9;
  int mx = 1e9;
  for (int i = 1; i <= n; i++) {
    int a;
    cin >> a;
    if (M[a] == 0)
      M[a] = i;
    else {
      if (mx > a) {
        mx = a;
        res = i - M[a];
      } else if (mx == a) {
        res = min(res, i - M[a]);
      }
      M[a] = i;
    }
  }
  cout << res << endl;
}

### Prompt

In CPP, your task is to solve 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;
map<int, int> M;
int n;
int main() {
  cin >> n;
  int res = 1e9;
  int mx = 1e9;
  for (int i = 1; i <= n; i++) {
    int a;
    cin >> a;
    if (M[a] == 0)
      M[a] = i;
    else {
      if (mx > a) {
        mx = a;
        res = i - M[a];
      } else if (mx == a) {
        res = min(res, i - M[a]);
      }
      M[a] = i;
    }
  }
  cout << res << endl;
}

```