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.

Example

Input

1 2


Output

111
#include<bits/stdc++.h>

#define ll long long
#define fornum(A,B,C) for(A=B;A<C;++A)
#define pii pair<int,int>
#define pll pair<ll,ll>

using namespace std;

/////////////////////////////////////////////////////
ll N, M, H, W;
ll mk1[1 << 11][1 << 11], mk2[1 << 11][1 << 11];
ll anss[1 << 11][1 << 11];
ll i, j, k;

int main(){
    scanf("%lld%lld", &N,&M);
    ll nm = max(N, M);
    fornum(i,0,nm){
        ll h, w;
        fornum(h,0,1<<i){
            ll a = 0;
            fornum(w,0,1<<i){
                a ^= anss[h][w];
            }
            fornum(w,0,1<<i){
                ll b = a ^ mk1[h][w + (1 << i) + 1];
                //printf("%lld %lld %lld\n", h,w + (1 << i), anss[h][w + (1 << i)]);
                if(mk1[h+1][w+1]!=b){
                    //printf("%lld %lld\n", mk1[h + 1][w + 1], b);
                    anss[h][w + (1 << i)] = 1;
                }
                a ^= anss[h][w + (1 << i)];
                mk1[h + 1][w + (1 << i) + 1] = mk1[h + 1][w + 1];
                //printf("%lld\n", anss[h][w + (1 << i)]);
            }
        }
        fornum(w,0,1<<i){
            ll a = 0;
            fornum(h,0,1<<i){
                a ^= anss[h][w];
            }
            fornum(h,0,1<<i){
                ll b = a ^ mk1[h + (1 << i) + 1][w];
                if(mk1[h+1][w+1]!=b){
                    anss[h + (1 << i)][w] = 1;
                }
                a ^= anss[h + (1 << i)][w];
                mk1[h + (1 << i) + 1][w + 1] = mk1[h + 1][w + 1];
            }
        }
        fornum(h,0,1<<i){
            ll a = 0;
            fornum(w,0,1<<i){
                a ^= anss[h + (1 << i)][w];
            }
            fornum(w,0,1<<i){
                ll b = a ^ mk1[h + (1 << i)][w + (1 << i) + 1];
                if(mk1[h+1][w+1]==b){
                    anss[h + (1 << i)][w + (1 << i)] = 1;
                }
                a ^= anss[h + (1 << i)][w + (1 << i)];
                mk1[h + (1 << i) + 1][w + (1 << i) + 1] = !mk1[h + 1][w + 1];
            }
        }
    }
    fornum(i,1,1<<N){
        fornum(j,1,1<<M){
            printf("%lld", anss[i][j]);
        }
        printf("\n");
    }
    return 0;
}