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 <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <utility>

using namespace std;
typedef long long ll;
const int LEN = 100000;

struct fastio {
  int it, len;
  char s[LEN + 5];
  fastio() {
    it = len = 0;
  }
  char get() {
    if(it < len) return s[it++];
    it = 0, len = fread(s, 1, LEN, stdin);
    return len ? s[it++] : EOF;
  }
  bool notend() {
    char c;
    for (c = get();c == ' ' || c == '\n';c = get());
    if (it) it--;
    return c != EOF;
  }
  void put(char c) {
    if(it == LEN) fwrite(s,1,LEN,stdout), it = 0;
    s[it++] = c;
  }
  void flush() {
    fwrite(s, 1, it, stdout);
  }
}buff, bufo;
inline int getint() {
  char c; int res = 0, sig = 1;
  for (c = buff.get(); c < '0' || c > '9'; c = buff.get()) if(c == '-') sig = -1;
  for (;c >= '0' && c <= '9';c = buff.get()) res = res * 10 + (c - '0');
  return sig * res;
}
inline ll getll() {
  char c; ll res = 0, sig = 1;
  for (c = buff.get(); c < '0' || c > '9'; c = buff.get()) if (c == '-') sig = -1;
  for (; c >= '0' && c <= '9'; c = buff.get()) res = res * 10 + (c - '0');
  return sig * res;
}
inline void putint(int x, char suf) {
  if (!x) bufo.put('0');
  else {
    if (x < 0) bufo.put('-'), x = -x;
    int k = 0; char s[15];
    while (x) {
      s[++k] = x % 10 + '0';
      x /= 10;
    }
    for (; k; k--) bufo.put(s[k]);
  }
  bufo.put(suf);
}
inline void putll(ll x, char suf) {
  if (!x) bufo.put('0');
  else {
    if (x < 0) bufo.put('-'), x = -x;
    int k = 0; char s[25];
    while (x) {
      s[++k] = x % 10 + '0';
      x /= 10;
    }
    for (; k; k--) bufo.put(s[k]);
  }
  bufo.put(suf);
}
inline char get_char() {
  char c;
  for (c = buff.get(); c == ' ' || c == '\n'; c = buff.get());
  return c;
}
bool a[1035][1035];
void build(int n) {
  if (!n) a[0][0] = 0;
  else {
    build(n - 1);
    int hf = (1 << (n - 1)), w = (1 << n);
    for (int i = 0; i < hf; i++)
      for (int j = hf; j < w; j++)
        a[i][j] = a[i][w - 1 - j];
    for (int i = hf; i < w; i++)
      for (int j = 0; j < hf; j++)
        a[i][j] = a[w - 1 - i][j];
    for (int i = hf; i < w; i++)
      for (int j = hf; j < w; j++)
        a[i][j] = !a[i][w - 1 - j];
  }
}
int n, m, res[1035][1035];
int main() {
  build(10);
  n = getint(), m = getint();
  for (int i = 1; i < (1 << n); i++) {
    for (int j = 1; j < (1 << m); j++) {
      bufo.put('0' + (a[i][j] ^ a[i - 1][j] ^ a[i][j - 1] ^ a[i - 1][j - 1]));
    }
    bufo.put('\n');
  }
  bufo.flush();
  return 0;
}
