#include <bits/stdc++.h>
using namespace std;
long power(long a, long p) {
  long res = 1;
  while (p > 0) {
    if (p & 1) res = (res * a);
    a = (a * a);
    p >>= 1;
  }
  return res;
}
string to_binary(long num) {
  string s;
  while (num) s.push_back(char(num & 1) + '0'), num >>= 1;
  reverse(s.begin(), s.end());
  return s;
}
long to_num(string num) {
  long ans = 0;
  for (auto& it : (num)) ans = ans * 2 + (it - '0');
  return ans;
}
void func() {
  int n;
  cin >> n;
  int arr[n + 1];
  int val = INT_MAX;
  for (int i = 1; i <= (n); i++) cin >> arr[i], val = min(val, arr[i]);
  int pos = n + 1;
  int ans = n + 1;
  for (int i = 1; i <= (n); i++) {
    if (arr[i] == val) {
      if (pos == n + 1)
        pos = i;
      else
        ans = min(ans, i - pos), pos = i;
    }
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  func();
  return 0;
}

### Prompt

Please create a solution in cpp 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;
long power(long a, long p) {
  long res = 1;
  while (p > 0) {
    if (p & 1) res = (res * a);
    a = (a * a);
    p >>= 1;
  }
  return res;
}
string to_binary(long num) {
  string s;
  while (num) s.push_back(char(num & 1) + '0'), num >>= 1;
  reverse(s.begin(), s.end());
  return s;
}
long to_num(string num) {
  long ans = 0;
  for (auto& it : (num)) ans = ans * 2 + (it - '0');
  return ans;
}
void func() {
  int n;
  cin >> n;
  int arr[n + 1];
  int val = INT_MAX;
  for (int i = 1; i <= (n); i++) cin >> arr[i], val = min(val, arr[i]);
  int pos = n + 1;
  int ans = n + 1;
  for (int i = 1; i <= (n); i++) {
    if (arr[i] == val) {
      if (pos == n + 1)
        pos = i;
      else
        ans = min(ans, i - pos), pos = i;
    }
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  func();
  return 0;
}

```