#include <bits/stdc++.h>
using namespace std;
const int M = 1000000001;
int main() {
  vector<int> vec;
  int n, a[200001];
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  int mi = *min_element(a, a + n);
  for (int i = 0; i < n; i++) {
    if (a[i] == mi) {
      vec.push_back(i);
    }
  }
  int64_t m = M;
  for (int i = 1; i < vec.size(); i++) {
    if ((vec[i] - vec[i - 1]) < m) {
      m = vec[i] - vec[i - 1];
    }
  }
  cout << m;
  return 0;
}

### 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;
const int M = 1000000001;
int main() {
  vector<int> vec;
  int n, a[200001];
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  int mi = *min_element(a, a + n);
  for (int i = 0; i < n; i++) {
    if (a[i] == mi) {
      vec.push_back(i);
    }
  }
  int64_t m = M;
  for (int i = 1; i < vec.size(); i++) {
    if ((vec[i] - vec[i - 1]) < m) {
      m = vec[i] - vec[i - 1];
    }
  }
  cout << m;
  return 0;
}

```