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. $\mathrm{2n2n people\; come\; to\; the\; event\; in nn pairs\; (couples).\; After\; a\; night\; of\; fun,\; Allen\; wants\; to\; line\; everyone\; up\; for\; a\; final\; picture.\; The 2n2n people\; line\; up,\; but\; Allen\; doesn\text{'}t\; like\; the\; ordering.\; Allen\; prefers\; if\; each\; pair\; occupies\; adjacent\; positions\; in\; the\; line,\; as\; this\; makes\; the\; picture\; more\; aesthetic.}$

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 $\mathrm{nn (1\le n\le 1001\le n\le 100),\; the\; number\; of\; pairs\; of\; people.}$

The second line contains $\mathrm{2n2n integers a1,a2,\dots ,a2na1,a2,\dots ,a2n.\; For\; each ii with 1\le i\le n1\le i\le n, ii appears\; exactly\; twice.\; If aj=ak=iaj=ak=i,\; that\; means\; that\; the jj-th\; and kk-th\; people\; in\; the\; line\; form\; a\; couple.}$

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 $\mathrm{11233244\to 11232344\to 1122334411233244\to 11232344\to 11223344 in\; two\; steps.\; Note\; that\; the\; sequence 11233244\to 11323244\to 1133224411233244\to 11323244\to 11332244 also\; works\; in\; the\; same\; number\; of\; steps.}$

The second sample case already satisfies the constraints; therefore we need $\mathrm{00 swaps.}$

#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("
");

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