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;
using lint = long long;
const lint mod = 998244353;
#define all(x) (x).begin(), (x).end()
#define bitcount(n) __builtin_popcountl((lint)(n))
#define fcout cout << fixed << setprecision(15)
#define highest(x) (63 - __builtin_clzl(x))
template<class T> inline void YES(T condition){ if(condition) cout << "YES" << endl; else cout << "NO" << endl; }
template<class T> inline void Yes(T condition){ if(condition) cout << "Yes" << endl; else cout << "No" << endl; }
template<class T = string, class U = char>int character_count(T text, U character){ int ans = 0; for(U i: text){ ans += (i == character); } return ans; }
lint power(lint base, lint exponent, lint module){ if(exponent % 2){ return power(base, exponent - 1, module) * base % module; }else if(exponent){ lint root_ans = power(base, exponent / 2, module); return root_ans * root_ans % module; }else{ return 1; }}
struct position{ int y, x; }; position mv[4] = {{0, -1}, {1, 0}, {0, 1}, {-1, 0}}; // double euclidean(position first, position second){ return sqrt((second.x - first.x) * (second.x - first.x) + (second.y - first.y) * (second.y - first.y)); }
template<class T, class U> string to_string(pair<T, U> x){ return to_string(x.first) + "," + to_string(x.second); } string to_string(string x){ return x; }
template<class itr> void array_output(itr start, itr goal){ string ans; for(auto i = start; i != goal; i++) ans += to_string(*i) + " "; if(!ans.empty()) ans.pop_back(); cout << ans << endl; }
template<class itr> void cins(itr first, itr last){ for(auto i = first; i != last; i++){ cin >> (*i); } }
template<class T> T gcd_calc(T a, T b){ if(b){ return gcd_calc(b, a % b); }else{ return a; }}
template<class T> T gcd(T a, T b){ if(a < b) swap(a, b); return gcd_calc(a, b); }
template<class T> T lcm(T a, T b){ return a / gcd(a, b) * b; }
struct combination{ vector<lint> fact, inv; combination(int sz) : fact(sz + 1), inv(sz + 1){ fact[0] = 1; for(int i = 1; i <= sz; i++){ fact[i] = fact[i - 1] * i % mod; } inv[sz] = power(fact[sz], mod - 2, mod); for(int i = sz - 1; i >= 0; i--){ inv[i] = inv[i + 1] * (i + 1) % mod; } } lint C(int p, int q) const{ if(q < 0 || p < q) return 0; return (fact[p] * inv[q] % mod * inv[p - q] % mod); } };
template<class itr> bool next_sequence(itr first, itr last, int max_bound){ itr now = last; while(now != first){ now--; (*now)++; if((*now) == max_bound){ (*now) = 0; }else{ return true; } } return false; }
inline int at(int x, int k){ return (x >> k) & 1; }

int arr[12][1030][1030];

int main(){
    int N, M;
    cin >> N >> M;
    int s = 10;
    arr[0][0][0] = 0;
    for(int i = 1; i <= s; i++){
        for(int j = 0; j < (1 << (i - 1)); j++){
            for(int k = 0; k < (1 << i); k++){
                arr[i][j][k] = arr[i - 1][j][k / 2];
            }
        }
        for(int j = 0; j < (1 << (i - 1)); j++){
            for(int k = 0; k < (1 << i); k += 2){
                arr[i][j + (1 << (i - 1))][k] = arr[i - 1][j][k / 2];
                arr[i][j + (1 << (i - 1))][k + 1] = !arr[i - 1][j][k / 2];
            }
        }
    }
    int m = max(N, M);
    int sum[1 << m][1 << m];
    for(int i = 0; i < (1 << m); i++){
        for(int j = 0; j < (1 << m); j++){
            sum[i][j] = arr[m][i][j] ^ arr[m][i][0];
            //cout << sum[i][j];
        }
        //cout << endl;
    }
    int ans[(1 << m) - 1][(1 << m) - 1];
    for(int i = 0; i < (1 << m) - 1; i++){
        for(int j = 0; j < (1 << m) - 1; j++){
            ans[i][j] = (sum[i][j] ^ sum[i + 1][j] ^ sum[i][j + 1] ^ sum[i + 1][j + 1]);
        }
    }
    if(N <= M){
        for(int i = 0; i < (1 << N) - 1; i++){
            for(int j = 0; j < (1 << M) - 1; j++){
                cout << ans[i][j];
            }
            cout << endl;
        }
    }else{
        for(int i = 0; i < (1 << N) - 1; i++){
            for(int j = 0; j < (1 << M) - 1; j++){
                cout << ans[j][i];
            }
            cout << endl;
        }
    }
}
