Gym - 101059B Shift and Push

Posted 2020-10-09 starry

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。

 1 #include <iostream>
 2 #include <stdio.h>
 3 #define ll long long
 4 #define INF 0x3f3f3f3f
 5 using namespace std;
 6 const int N = 1e6+10;
 7 int a[N], b[N/10+10][5];
 8 int main() {
 9     int n;
10     cin >> n;
11     for(int i = 1; i <= n; i ++){
12         scanf("%d", &a[i]);
13         if(!b[a[i]][0]) b[a[i]][1] = b[a[i]][2] = b[a[i]][3] = i;
14         else {
15             b[a[i]][2] = b[a[i]][3];
16             b[a[i]][3] = i;
17             b[a[i]][4] = max(b[a[i]][4], b[a[i]][3] - b[a[i]][2] - 1);            
18         } 
19         b[a[i]][0] ++;
20     }
21     int ans = INF;
22     for(int i = 1; i <= n; i ++) {
23         b[a[i]][4] = max(b[a[i]][4], n - b[a[i]][3] + b[a[i]][1] - 1);
24         ans = min(ans, b[a[i]][4] + n);
25     }
26     printf("%d\n",ans);
27     return 0;
28 }