B. Finding Team Member

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There is a programing contest named SnakeUp, 2n people want to compete for it. In order to attend this contest, people need to form teams of exactly two people. You are given the strength of each possible combination of two people. All the values of the strengths are distinct.

Every contestant hopes that he can find a teammate so that their team’s strength is as high as possible. That is, a contestant will form a team with highest strength possible by choosing a teammate from ones who are willing to be a teammate with him/her. More formally, two people A and B may form a team if each of them is the best possible teammate (among the contestants that remain unpaired) for the other one.

Can you determine who will be each person’s teammate?

Input

There are 2n lines in the input.

The first line contains an integer n (1 ≤ n ≤ 400) — the number of teams to be formed.

The i-th line (i > 1) contains i - 1 numbers ai1, ai2, ... , ai(i - 1). Here aij (1 ≤ aij ≤ 106, all aij are distinct) denotes the strength of a team consisting of person i and person j (people are numbered starting from 1.)

Output

Output a line containing 2n numbers. The i-th number should represent the number of teammate of i-th person.

Examples

input

261 23 4 5

output

2 1 4 3

input

34870603831 161856845957 794650 97697783847 50566 691206 498447698377 156232 59015 382455 626960

output

6 5 4 3 2 1

Note

In the first sample, contestant 1 and 2 will be teammates and so do contestant 3 and 4, so the teammate of contestant 1, 2, 3, 4 will be 2, 1, 4, 3 respectively.

解题说明：题目的意思是每两个人搭档（就是每组组队可能）都对应一个权值，我们不要求全局最优 只希望n次组队中的每次，都使得最强的队伍尽可能强， 要你输出匹配情况。

#include<cstdio>#include <cstring>#include<cmath>#include<iostream>#include<algorithm>#include<vector>#include <map>using namespace std;int a[1000005][2],ans[1000];int main(){int n,i,j,x;scanf("%d",&n);for( i=1;i<=2*n;i++){for(j=1;j<i;j++){scanf("%d",&x);a[x][0]=i;a[x][1]=j;}}for(i=1000000;i>0;i--){if(ans[a[i][0]]==0 && ans[a[i][1]]==0){ans[a[i][0]]=a[i][1];ans[a[i][1]]=a[i][0];}}for(int i=1;i<=2*n;i++){printf("%d ",ans[i]);}printf("\n");return 0;}