#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  long long int mnm = 1000000005;
  vector<long long int> v(n);
  for (int i = 0; i < n; ++i) {
    cin >> v[i];
    if (v[i] < mnm) {
      mnm = v[i];
    }
  }
  vector<int> dis;
  for (int i = 0; i < n; ++i) {
    if (v[i] == mnm) {
      dis.push_back(i);
    }
  }
  int ans = 100005;
  for (int i = 0; i < dis.size() - 1; ++i) {
    if (dis[i + 1] - dis[i] < ans) {
      ans = dis[i + 1] - dis[i];
    }
  }
  cout << ans;
}

### 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;
  cin >> n;
  long long int mnm = 1000000005;
  vector<long long int> v(n);
  for (int i = 0; i < n; ++i) {
    cin >> v[i];
    if (v[i] < mnm) {
      mnm = v[i];
    }
  }
  vector<int> dis;
  for (int i = 0; i < n; ++i) {
    if (v[i] == mnm) {
      dis.push_back(i);
    }
  }
  int ans = 100005;
  for (int i = 0; i < dis.size() - 1; ++i) {
    if (dis[i + 1] - dis[i] < ans) {
      ans = dis[i + 1] - dis[i];
    }
  }
  cout << ans;
}

```