/* Linear time (in the size of the output) solution for routing.
 * David Garcia Soriano */
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;

const int maxn = 13;

int perm[1 << maxn], inv[1 << maxn];           // current permutation, inverse permutation
char switched[2 * maxn - 1][1 << (maxn - 1)],  // state of switches
     path[1 << maxn];                          // 0 for left path, 1 for right

void dfs(int v, int r) {
  path[v] = r;
  if (path[v ^ 1] < 0) dfs(v ^ 1, r ^ 1);
  v = perm[inv[v] ^ 1];
  if (path[v] < 0) dfs(v, r ^ 1);
}

int perfectshuffle(int i, int n) { return ((i & 1) << (n - 1)) | (i >> 1); }

/* Solve the routing problem with n nodes, where src[i] is matched to dest[i]
 * and switch submatrix starting at (ypos, xpos), and store the state of the switches. */
void route_all(int n, int ypos, int xpos, const vector<int>& src, const vector<int>& dest) {
  if (n == 1) { switched[ypos][xpos] = src[0] != dest[0]; return; }
  
  int xlen = 1 << n, ylen = 2 * n - 1;

  /* Write dest as a permutation of src and compute the inverse.
   * From now on we pretend the values of src are 0, 1, .., xlen -1. */
  for (int i = 0; i < xlen; ++i) inv[src[i]] = i;
  for (int i = 0; i < xlen; ++i) perm[i] = inv[dest[i]];
  for (int i = 0; i < xlen; ++i) inv[perm[i]] = i;
 
  /* Find a bicoloring (to select which messages will go through the left half of the network).
   * The order is chosen so that low-numbered switches are turned off if possible. */
  memset(path, -1, sizeof(path));
  for (int i = 0; i < xlen; ++i) if (path[i] < 0) dfs(i, 0);

  // Compute new src/dest pairs (1 is the upper part, 2 the lower part).
  vector<int> left1, right1;
  for (int i = 0; i < xlen; i += 2) { // upper part
    switched[ypos][xpos + i / 2] = path[i];
    for (int j = 0; j < 2; ++j) {
      int x = perfectshuffle((i | j) ^ path[i], n);
      if (x < xlen / 2) left1.push_back(i | j); 
      else right1.push_back(i | j);
    }
  }

  vector<int> left2(xlen / 2), right2(xlen / 2);
  for (int i = 0; i < xlen; i += 2) {   // lower part
    int s = switched[ypos + ylen - 1][xpos + i / 2] = path[perm[i]];
    for (int j = 0; j < 2; ++j) {
      int x = perfectshuffle((i | j) ^ s, n);
      if (x < xlen / 2) left2[x] = perm[i | j]; 
      else right2[x - xlen / 2] = perm[i | j];
    }
  }

  // Recurse on the left and right parts
  route_all(n - 1, ypos + 1, xpos, left1, left2);
  route_all(n - 1, ypos + 1, xpos + xlen / 4, right1, right2);
}

int main() {
  int n;
  bool first = true;
  while (scanf("%i", &n) == 1, n) {
    int xlen = 1 << n, ylen = 2 * n - 1;
    vector<int> dest(xlen), src(xlen);
    for (int i = 0; i < xlen; ++i) src[i] = i;
    for (int i = 0; i < xlen; ++i) scanf("%i", &dest[i]);

    route_all(n, 0, 0, src, dest);

    if (!first) puts("");
    else first = false;

    for (int y = 0; y < ylen; ++y) {
      for (int x = 0; x < xlen / 2; ++x)
        putchar('0' + switched[y][x]);
      puts("");
    }
  }
  return 0;
}