B. Shift and Push

B. Shift and Push

time limit per test

1.0 s

memory limit per test

256 MB

input

standard input

output

standard output

Given two equally sized arrays A and B of size N. A is empty and B has some values.

You need to fill A with an element X such that X belongs to B.

The only operations allowed are:

1. Copy B[i] to A[i].

2. Cyclic shift B by 1 to the the right.

You need to minimise the number of operations.

Input

The first line contains a single positive integer N(1 ≤ N ≤ 10⁶), denoting the size of the arrays.

Next line contains N space separated positive integers denoting the elements of the array B(1 ≤ B[i] ≤ 10⁵).

Output

Output a single integer, denoting the minimum number of operations required.

Examples

Input

3
1 2 3

Output

5

Input

6
1 1 2 2 3 3

Output

10

Note

In the first test case:

We can have 5 steps as: fill first element, shift, fill second element, shift, fill third element.

Initially, A = [_, _, _], B = [1, 2, 3]

After step 1, A = [1, _, _], B = [1, 2, 3]

After step 2, A = [1, _, _], B = [3, 1, 2]

After step 3, A = [1, 1, _], B = [3, 1, 2]

After step 4, A = [1, 1, _], B = [2, 3, 1]

After step 5, A = [1, 1, 1], B = [2, 3, 1]

求每个数中相邻区间最大的最小的那个数。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#define ll long long int
#define inf 0x3f3f3f3f
#define N 1000005
#define mem(a) memset(a,0,sizeof(a))
using namespace std;
int lowbit(int x){ return x&(-x);}
int a[N],b[N/10+5][5];

int main(){
  int n;
  cin>>n;
  for(int i = 1; i <= n; i ++){
      scanf("%d", &a[i]);
      if(!b[a[i]][0]) b[a[i]][1] = b[a[i]][2] = b[a[i]][3] = i;
      else {
          b[a[i]][2] = b[a[i]][3];
          b[a[i]][3] = i;
          b[a[i]][4] = max(b[a[i]][4], b[a[i]][3] - b[a[i]][2] - 1);
      }
      b[a[i]][0] ++;
  }
  int ans = inf;
  for(int i = 1; i <= n; i ++) {
      b[a[i]][4] = max(b[a[i]][4], n - b[a[i]][3] + b[a[i]][1] - 1);
      ans = min(ans, b[a[i]][4] + n);
  }
  printf("%d
",ans);
  return 0;
}