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;
void files() {
  freopen("1.in", "r", stdin);
  freopen("1.out", "w", stdout);
  return;
}
double eps = 1e-6;
double pi = acos(-1);
const int MXN = 1e5 + 5;
const long long MOD = 1e9 + 7;
int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1, -1, 1, -1, 0, 0, -1, 1};
int dx4[] = {1, -1, 0, 0};
int dy4[] = {0, 0, 1, -1};
long long bin(long long base, long long pow, long long m) {
  base %= m;
  long long ans = 1;
  while (pow) {
    if (pow & 1) ans = ans * base % m;
    base = base * base % m;
    pow >>= 1;
  }
  return ans;
}
void sol() {
  int n;
  scanf("%d", &n);
  map<int, vector<int>> id;
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%d", &x);
    id[x].push_back(i);
  }
  int mn = n;
  for (auto v : id) {
    for (int i = 0; i + 1 < (int)v.second.size(); ++i) {
      mn = min(mn, v.second[i + 1] - v.second[i]);
    }
    break;
  }
  printf("%d\n", mn);
}
int main() {
  int tcase = 1;
  cout << fixed << setprecision(6);
  while (tcase--) {
    sol();
  }
}
