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;
const double PI = acos(-1);
const double eps = 1e-8;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e5 + 6;
long long qpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return ans;
}
long long inv(long long a) { return qpow(a, MOD - 2); }
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
int a[MAXN], b[MAXN], n;
vector<int> s;
int main() {
  while (cin >> n) {
    for (int i = 1; i <= n; i++) cin >> a[i], b[i] = a[i];
    sort(a + 1, a + 1 + n);
    int minn = a[1];
    for (int i = 1; i <= n; i++) {
      if (b[i] == a[1]) s.push_back(i);
    }
    int ans = INF;
    int len = s.size();
    for (int i = 0; i < len - 1; i++) {
      ans = min(ans, s[i + 1] - s[i]);
    }
    cout << ans << endl;
  }
  return 0;
}
