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>
using namespace std;
#define int long long
#define ld long double
#define ll long long
#define P pair<int,int>
#define fi first
#define se second
#define rep(i,n) for(int i=0;i<n;i++)
#define all(v) v.begin(),v.end()
#define pb push_back
#define eb emplace_back
template<class T>void chmax(T &a,T b){if(a<b)a=b;}
template<class T>void chmin(T &a,T b){if(a>b)a=b;}
constexpr int INF=1000000000000000000;
constexpr int mod=998244353;
constexpr double eps=1e-8;
const double pi=acos(-1);
int dx[]={0,1,0,-1},dy[]={1,0,-1,0};
int Random(int mi,int ma){
    random_device rnd;
    mt19937 mt(rnd());//32bit
    //[mi,ma]
    uniform_int_distribution<int>engine(mi,ma);
    return engine(mt);
}
int gcd(int a,int b){
    if(b==0)return a;
    return gcd(b,a%b);
}
int lcm(int a,int b){
    return a/gcd(a,b)*b;
}
bool prime(int a){
    if(a==1)return false;
    for(int i=2;i*i<=a;i++){
        if(a%i==0)return false;
    }
    return true;
}
int modpow(int a,int b){
    if(b==0)return 1;
    if(b%2)return modpow(a,b-1)*a%mod;
    int memo=modpow(a,b/2);
    return memo*memo%mod;
}
vector<int>kaijo,invkaijo;
void init_fact(int n){
    kaijo.resize(n+1);
    invkaijo.resize(n+1);
    kaijo[0]=1;
    for(int i=1;i<=n;i++){
        kaijo[i]=kaijo[i-1]*i;
        kaijo[i]%=mod;
    }
    rep(i,n+1)invkaijo[i]=modpow(kaijo[i],mod-2);
}
int comb(int a,int b){
    if(a<b)return 0;
    if(a<0||b<0)return 0;
    return kaijo[a]*modpow(kaijo[a-b],mod-2)%mod*modpow(kaijo[b],mod-2)%mod;
}
int inv(int x){
    x=modpow(x,mod-2);
    return x;
}
int a[1<<12][1<<12],ans[1<<12][1<<12];
void calc(int k){
    if(k==0){
        a[0][0]=0;
        return;
    }
    calc(k-1);
    int n=1<<(k-1);
    rep(i,n){
        rep(j,n){
            a[i+n][j]=a[i][j];
            a[i][j+n]=a[i][j];
            a[i+n][j+n]=1-a[i][j];
        }
    }
}
int n,m;
signed main(){
    cin.tie(0);ios::sync_with_stdio(false);
    cin>>n>>m;
    int N=max(n,m);
    calc(N);
    N=1<<N;
    rep(i,N-1){
        rep(j,N-1){
            if((a[i][j+1]+a[i+1][j]+a[i][j])%2==a[i+1][j+1]%2)ans[i][j]=0;
            else ans[i][j]=1;
        }
    }
    rep(i,(1<<n)-1){
        rep(j,(1<<m)-1){
            cout<<ans[i][j];
        }
        cout<<endl;
    }
    return 0;
}