Problem:
Score : 900 points

Problem Statement

We have a grid with (2^N - 1) rows and (2^M-1) columns.
You are asked to write 0 or 1 in each of these squares.
Let a_{i,j} be the number written in the square at the i-th row from the top and the j-th column from the left.

For a quadruple of integers (i_1, i_2, j_1, j_2) such that 1\leq i_1 \leq i_2\leq 2^N-1, 1\leq j_1 \leq j_2\leq 2^M-1, let
S(i_1, i_2, j_1, j_2) = \displaystyle \sum_{r=i_1}^{i_2}\sum_{c=j_1}^{j_2}a_{r,c}.
Then, let the oddness of the grid be the number of quadruples (i_1, i_2, j_1, j_2) such that S(i_1, i_2, j_1, j_2) is odd.

Find a way to fill in the grid that maximizes its oddness.

Constraints

N and M are integers between 1 and 10 (inclusive).

Input

Input is given from Standard Input in the following format:

N M

Output

Print numbers to write in the grid so that its oddness is maximized, in the following format:

a_{1,1}a_{1,2}\cdots a_{1,2^M-1}
a_{2,1}a_{2,2}\cdots a_{2,2^M-1}
\vdots
a_{2^N-1,1}a_{2^N-1,2}\cdots a_{2^N-1,2^M-1}

If there are multiple solutions, you can print any of them.

Sample Input 1

1 2

Sample Output 1

111

For this grid, S(1, 1, 1, 1), S(1, 1, 2, 2), S(1, 1, 3, 3), and S(1, 1, 1, 3) are odd, so it has the oddness of 4.

We cannot make the oddness 5 or higher, so this is one of the ways that maximize the oddness.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define ll long long
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define repp(i,l,r,k)for(ll i=(l);i<(r);i+=(k))
#define INF ((1LL<<62)-(1LL<<31))
#define max(p,q)((p)>(q)?(p):(q))
#define min(p,q)((p)<(q)?(p):(q))
#define bit(n,m)(((n)>>(m))&1)
int upll(const void*a, const void*b){return*(ll*)a<*(ll*)b?-1:*(ll*)a>*(ll*)b?1:0;}
int downll(const void*a, const void*b){return*(ll*)a<*(ll*)b?1:*(ll*)a>*(ll*)b?-1:0;}
void sortup(ll*a,int n){qsort(a,n,sizeof(ll),upll);}
void sortdown(ll*a,int n){qsort(a,n,sizeof(ll),downll);}
ll pom(ll a,ll n,int m){ll x=1;for(a%=m;n;n/=2)n&1?x=x*a%m:0,a=a*a%m;return x;}
//#define MOD 998244353
#define MOD 1000000007
#define invp(a,p)pom(a,p-2,p)


//辺の情報を個別に持つタイプ
typedef struct edge{ll s,g,c;}E;
typedef struct graph{
	int vcnt,ecnt;
	E  e[400010];//適宜変える(ecnt)
	int id[200010];//適宜変える(vcnt)
}G;
G g;

int esort(const void*a,const void*b){
	E*p=(E*)a,*q=(E*)b;
	if((*p).s<(*q).s)return -1;
	if((*p).s>(*q).s)return  1;
	if((*p).g<(*q).g)return -1;
	return 1;
}
void readgraph(){
	//適宜変える
	ll n;
	scanf("%lld",&n);
	rep(i,0,n-1){
		ll x,y;
		scanf("%lld%lld",&x,&y);
		x--,y--;
		g.e[2*i].s=x;
		g.e[2*i].g=y;
		g.e[2*i+1].s=y;
		g.e[2*i+1].g=x;
	}
	g.vcnt=n;
	g.ecnt=2*n-2;
	qsort(g.e,g.ecnt,sizeof(E),esort);

	int p=0;
	rep(i,0,g.vcnt){
		while(p<g.ecnt&&g.e[p].s<i)p++;
		g.id[i]=p;
	}
	g.id[g.vcnt]=g.ecnt;//番兵
}


ll dp[200010];

ll dfs(int v,int pre,int c){
    dp[v]=c;
	rep(i,g.id[v],g.id[v+1]){
		if(g.e[i].g!=pre){
			dfs(g.e[i].g,v,c^1);
		}
	}
}

int main(){
    readgraph();
    dfs(0,-1,0);
    
    
    int n1=(g.vcnt+2)/3;
    int n2=(g.vcnt+1)/3;
    int n3=g.vcnt/3;
    
    int cnt1=0;
    rep(i,0,g.vcnt)if(dp[i])cnt1++;
    if(cnt1<=n3){
        rep(i,0,g.vcnt){
            if(dp[i]){
                printf("%d ",n3--*3);
            }else{
                if(n1)printf("%d ",n1--*3-2);
                else if(n2)printf("%d ",n2--*3-1);
                else printf("%d ",n3--*3);
            }
        }
    }else if(g.vcnt-cnt1<=n3){
        rep(i,0,g.vcnt){
            if(!dp[i]){
                printf("%d ",n3--*3);
            }else{
                if(n1)printf("%d ",n1--*3-2);
                else if(n2)printf("%d ",n2--*3-1);
                else printf("%d ",n3--*3);
            }
        }
    }else{
        rep(i,0,g.vcnt){
            if(dp[i]){
                if(n1)printf("%d ",n1--*3-2);
                else printf("%d ",n3--*3);
            }else{
                if(n2)printf("%d ",n2--*3-1);
                else printf("%d ",n3--*3);
            }
        }
    }
}