Problem: During the break, we decided to relax and play dominoes. Our box with Domino was empty, so we decided to borrow the teacher's dominoes.

The teacher responded instantly at our request. He put nm dominoes on the table as an n × 2m rectangle so that each of the n rows contained m dominoes arranged horizontally. Each half of each domino contained number (0 or 1).

We were taken aback, and the teacher smiled and said: "Consider some arrangement of dominoes in an n × 2m matrix. Let's count for each column of the matrix the sum of numbers in this column. Then among all such sums find the maximum one. Can you rearrange the dominoes in the matrix in such a way that the maximum sum will be minimum possible? Note that it is prohibited to change the orientation of the dominoes, they all need to stay horizontal, nevertheless dominoes are allowed to rotate by 180 degrees. As a reward I will give you all my dominoes".

We got even more taken aback. And while we are wondering what was going on, help us make an optimal matrix of dominoes.

Input

The first line contains integers n, m (1 ≤ n, m ≤ 103).

In the next lines there is a description of the teachers' matrix. Each of next n lines contains m dominoes. The description of one domino is two integers (0 or 1), written without a space — the digits on the left and right half of the domino.

Output

Print the resulting matrix of dominoes in the format: n lines, each of them contains m space-separated dominoes.

If there are multiple optimal solutions, print any of them.

Examples

Input

2 3
01 11 00
00 01 11


Output

11 11 10
00 00 01


Input

4 1
11
10
01
00


Output

11
10
01
00

Note

Consider the answer for the first sample. There, the maximum sum among all columns equals 1 (the number of columns is 6, and not 3). Obviously, this maximum can't be less than 1, then such matrix is optimal.

Note that the dominoes can be rotated by 180 degrees.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct edje {
  int from, to;
  long long w;
  edje(int from, int to, long long w) : from(from), to(to), w(w) {}
  bool operator<(const edje& e) const { return w < e.w; }
};
const long long oo = (long long)((1e18) + 1);
const double eps = 1e-9;
const long long mod7 = 1000000007;
long long mod(long long a, long long b) { return (a % b + b) % b; }
long long strTonumModa(string s, long long athemod) {
  long long ans = 0;
  for (long long i = 0; i < ((long long)s.size()); i++) {
    ans = (ans * 10 + s[i] - '0') % athemod;
  }
  return ans;
}
long long mod_exp(long long base, long long p, long long MOD) {
  if (p == 0) return 1;
  if (p == 1) return base;
  long long ret = mod_exp(base, p / 2, MOD) % MOD;
  ret = (ret * ret) % MOD;
  if (p % 2 == 1) ret = (ret * base) % MOD;
  return ret;
}
string getstr(long long x) {
  stringstream ss;
  ss << x;
  string str = ss.str();
  return str;
}
bool comp(pair<long long, pair<long long, long long> > a,
          pair<long long, pair<long long, long long> > b) {
  return a.first > b.first;
}
int dr[] = {1, 1, 0, -1, -1, -1, 0, 1};
int dc[] = {0, 1, 1, 1, 0, -1, -1, -1};
long long sq(long long n) { return n * n; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<string> ans, in(n * m);
  for (int i = 0; i < (int)(n * m); ++i) cin >> in[i];
  sort(in.begin(), in.end(), greater<string>());
  bool f = true;
  string best[n][m];
  int r = 0;
  for (int i = 0; i < n * m; i += m) {
    vector<string> temp;
    if (f) {
      for (int j = (i); j < (int)(i + (m - 1) + 1); ++j) temp.push_back(in[j]);
    } else {
      for (int j = (i + (m - 1)); j >= (int)(i); --j) temp.push_back(in[j]);
    }
    for (int t = 0; t < ((long long)temp.size()); ++t) {
      best[r][t] += temp[t];
    }
    r++;
    f = !f;
  }
  for (int i = (1); i < (int)(n); ++i) {
    for (int j = (0); j < (int)(m); ++j) {
      if (best[i - 1][j] == "10" && best[i][j] == "10") {
        best[i][j] = "01";
      } else if (best[i - 1][j] == "01" && best[i][j] == "01") {
        best[i][j] = "10";
      }
    }
  }
  for (int i = 0; i < (int)(n); ++i) {
    for (int j = 0; j < (int)(m); ++j) {
      cout << best[i][j] << " ";
    }
    cout << endl;
  }
  return 0;
}