code forces 996D Suit and Tie

D. Suit and Tie

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Allen is hosting a formal dinner party. 2n people line up, but Allen doesn't like the ordering. Allen prefers if each pair occupies adjacent positions in the line, as this makes the picture more aesthetic.

Help Allen find the minimum number of swaps of adjacent positions he must perform to make it so that each couple occupies adjacent positions in the line.

Input

The first line contains a single integer 1≤n≤100), the number of pairs of people.

The second line contains k-th people in the line form a couple.

Output

Output a single integer, representing the minimum number of adjacent swaps needed to line the people up so that each pair occupies adjacent positions.

Examples

input

Copy

4
1 1 2 3 3 2 4 4

output

Copy

2

input

Copy

3
1 1 2 2 3 3

output

Copy

0

input

Copy

3
3 1 2 3 1 2

output

Copy

3

Note

In the first sample case, we can transform 11233244→11323244→11332244 also works in the same number of steps.

The second sample case already satisfies the constraints; therefore we need 0 swaps.

 

#include<bits/stdc++.h>
using namespace std;
int a[205];
int main(){
    int n;
    while(~scanf("%d",&n)){
        for(int i=0;i<2*n;i++){
            scanf("%d",&a[i]);
        }
        int ans=0;
        int pos;
        for(int i=1;i<2*n;i+=2){
            if(a[i]!=a[i-1]){
                int t=a[i];
               for(int j=i+1;j<2*n;j++){
                    if(a[j]==a[i-1]){
                        ans+=j-i;
                        pos=j;
                        a[i]=a[j];
                        break;
                    }
               }
               //这个就是那一段
               for(int j=pos;j>i;j--){
                    a[j]=a[j-1];
               }
                a[i+1]=t;

            }
//            for(int j=0;j<2*n;j++){
//                printf("%d ",a[j]);
//            }
//            printf("\n");

        }
        printf("%d\n",ans);
    }
    return 0;
}