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;
typedef long long ll;
typedef pair<ll, ll> l_l;
typedef pair<int, int> i_i;
template<class T>
inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}

const long double EPS = 1e-10;
const long long INF = 1e18;
const long double PI = acos(-1.0L);
//const ll mod = 1000000007;
vector<vector<vector<int>>> dp;

void initialize() {
    dp.push_back(vector<vector<int>>());
    dp[0].push_back({1});
    for(int h = 1; h <= 10; h++) {
        dp.push_back(vector<vector<int>>());
        for(int i = 0; i < dp[h-1].size(); i++) {
            dp[h].push_back({});
            for(auto val : dp[h-1][i]) {
                dp[h].back().push_back(val);
                dp[h].back().push_back(val);
            }
            dp[h].push_back({});
            for(auto val : dp[h-1][i]) {
                dp[h].back().push_back(val);
                dp[h].back().push_back(val ^ 1);
            }
        }
    }
}

void print(int idx) {
    for(auto tmp : dp[idx]) {
        for(auto tmp2 : tmp) {
            cerr << tmp2 << " ";
        }
        cerr << endl;
    }
    cerr << "end: " << idx << endl;
}

bool SWAP = false;
vector<vector<int>> ans;

void printans() {
    if(!SWAP) {
        for(auto tmp : ans) {
            for(int i = 0; i < tmp.size(); i++) {
                cout << tmp[i];
            }
            cout << endl;
        }
        /*
        cout << endl;
        for(int h = 0; h < ans.size(); h++) {
            int now = 0;
            cout << now;
            for(int w = 0; w < ans[0].size(); w++) {
                now ^= ans[h][w];
                cout << now;
            }
            cout << endl;
        }
        */
        return;
    }
    for(int i = 0; i < ans[0].size(); i++) {
        for(int j = 0; j < ans.size(); j++) {
            cout << ans[j][i];
        }
        cout << endl;
    }
}

int main() {
    //cout.precision(10);
    cin.tie(0);
    ios::sync_with_stdio(false);
    initialize();
    int N, M;
    cin >> N >> M;
    int H = pow(2, N) - 1;
    int W = pow(2, M) - 1;
    print(0);
    print(1);
    print(2);
    print(3);
    if(H > W) {
        swap(H, W);
        swap(N, M);
        SWAP = true;
    }
    ans = vector<vector<int>>(H, vector<int>(W));
    /*
    for(int h = 0; h < (H + 1) / 2; h++) {
        for(int w = 0; w < W; w++) {
            if(dp[M][h][w] != dp[M][h][w+1]) ans[h][w] = 1;
        }
    }
    for(int h = (H + 1) / 2; h < H; h++) {
        ans[h] = ans[H-1-h];
    }
    */
    for(int h = 0; h < H; h++) {
        for(int w = 0; w < W; w++) {
            ans[h][w] = dp[M][h+1][w+1] ^ dp[M][h+1][w] ^ dp[M][h][w+1] ^ dp[M][h][w];
        }
    }
    /*
    for(int h = 0; h < H; h++) {
        int now = 0;
        cout << now;
        for(int w = 0; w < W; w++) {
            now ^= ans[h][w];
            cout << " " << now;
        }
        cout << endl;
    }
    */
    printans();
    return 0;
}