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;
long long a, b, n, m, x, k, c, y, z, i, j, l, r, u, sum, maxx, v[2555][2555], d;
long long na, ma, nb, mb;
long long br[1456789], ar[1456789], mn[1456789], mx[1456789], used[1456789];
char h[1234][1234], p[55][55], g[123][123];
string s, t, ress;
vector<long long> vec, as;
map<pair<long long, long long>, long long> mmap;
set<long long> ss;
bool ok, oka[1456789];
char ch;
map<pair<int, int>, int> mpp;
void dii() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> ar[i];
  }
  x = *min_element(ar, ar + n);
  for (int i = 0; i < n; i++) {
    if (ar[i] == x) {
      a = b = i;
      break;
    }
  }
  sum = 2e9;
  for (long long i = a + 1; i < n; i++) {
    if (ar[i] == x) {
      sum = min(sum, i - b);
      b = i;
    }
  }
  cout << sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int q = 1;
  while (q--) dii();
  return 0;
}
