#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  vector<pair<long long int, int> > v;
  for (int i = 0; i < n; i++) {
    long long int a;
    cin >> a;
    v.push_back({a, i});
  }
  sort(v.begin(), v.end());
  int minn = 1000000;
  for (int i = 1; i < v.size(); i++) {
    if (v[i].first != v[i - 1].first)
      break;
    else {
      minn = min(minn, abs(v[i].second - v[i - 1].second));
    }
  }
  cout << minn;
}

### 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;
  vector<pair<long long int, int> > v;
  for (int i = 0; i < n; i++) {
    long long int a;
    cin >> a;
    v.push_back({a, i});
  }
  sort(v.begin(), v.end());
  int minn = 1000000;
  for (int i = 1; i < v.size(); i++) {
    if (v[i].first != v[i - 1].first)
      break;
    else {
      minn = min(minn, abs(v[i].second - v[i - 1].second));
    }
  }
  cout << minn;
}

```