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
#include <bits/stdc++.h>
using namespace std;
inline int scanInt() {
  int n = 0;
  char ch = getchar();
  int sign = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') sign = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    n = (n << 1) + (n << 3) + (int)(ch - '0');
    ch = getchar();
  }
  return n * sign;
}
inline long long int scanLong() {
  long long int n = 0;
  char ch = getchar();
  long long int sign = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') sign = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    n = (n << 1) + (n << 3) + (long long int)(ch - '0');
    ch = getchar();
  }
  return n * sign;
}
int main() {
  int n;
  cin >> n;
  int i;
  vector<int> v(n);
  for (i = 0; i < n; i++) {
    scanf("%d", &v[i]);
  }
  int ans = 100005;
  int m = *min_element(v.begin(), v.end());
  i = 0;
  while (v[i] != m) i++;
  int j = i + 1;
  while (j < n) {
    while (j < n && v[j] != m) j++;
    if (j != n) ans = min(ans, j - i);
    i = j;
    j++;
  }
  cout << ans;
  return 0;
}
