Problem: Let's define logical OR as an operation on two logical values (i. e. values that belong to the set {0, 1}) that is equal to 1 if either or both of the logical values is set to 1, otherwise it is 0. We can define logical OR of three or more logical values in the same manner:

<image> where <image> is equal to 1 if some ai = 1, otherwise it is equal to 0.

Nam has a matrix A consisting of m rows and n columns. The rows are numbered from 1 to m, columns are numbered from 1 to n. Element at row i (1 ≤ i ≤ m) and column j (1 ≤ j ≤ n) is denoted as Aij. All elements of A are either 0 or 1. From matrix A, Nam creates another matrix B of the same size using formula:

<image>.

(Bij is OR of all elements in row i and column j of matrix A)

Nam gives you matrix B and challenges you to guess matrix A. Although Nam is smart, he could probably make a mistake while calculating matrix B, since size of A can be large.

Input

The first line contains two integer m and n (1 ≤ m, n ≤ 100), number of rows and number of columns of matrices respectively.

The next m lines each contain n integers separated by spaces describing rows of matrix B (each element of B is either 0 or 1).

Output

In the first line, print "NO" if Nam has made a mistake when calculating B, otherwise print "YES". If the first line is "YES", then also print m rows consisting of n integers representing matrix A that can produce given matrix B. If there are several solutions print any one.

Examples

Input

2 2
1 0
0 0


Output

NO


Input

2 3
1 1 1
1 1 1


Output

YES
1 1 1
1 1 1


Input

2 3
0 1 0
1 1 1


Output

YES
0 0 0
0 1 0

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << endl;
  err(++it, args...);
}
const long long MOD = 1000000007LL, INF = 1e9, LINF = 1e18;
const long double Pi = 3.141592653589793116, EPS = 1e-9;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
const long long dx[4] = {0, 1, 0, -1};
const long long dy[4] = {1, 0, -1, 0};
void solve() {
  long long m, n;
  cin >> m >> n;
  long long arr[m][n];
  for (long long i = 0; i < m; i++)
    for (long long j = 0; j < n; j++) cin >> arr[i][j];
  long long ans[1000][1000] = {0};
  map<long long, map<long long, long long>> ha;
  for (long long i = 0; i < m; i++) {
    for (long long j = 0; j < n; j++) {
      if (!arr[i][j]) {
        for (long long k = 0; k < n; k++) ha[i][k] = 1;
        for (long long k = 0; k < m; k++) ha[k][j] = 1;
      }
    }
  }
  for (long long i = 0; i < m; i++) {
    for (long long j = 0; j < n; j++) {
      if (arr[i][j]) {
        long long flg = 0;
        for (long long k = 0; k < m; k++) {
          if (ans[k][j]) {
            flg = 1;
            break;
          }
          if (!ha[k][j]) {
            ans[k][j] = 1;
            ha[k][j] = 1;
            flg = 1;
            break;
          }
        }
        if (!flg) {
          for (long long k = 0; k < n; k++) {
            if (ans[i][k]) {
              flg = 1;
              break;
            }
            if (!ha[i][k]) {
              ans[i][k] = 1;
              ha[i][k] = 1;
              flg = 1;
              break;
            }
          }
        }
        if (!flg) {
          cout << "NO\n";
          return;
        }
      }
    }
  }
  cout << "YES\n";
  for (long long i = 0; i < m; i++) {
    for (long long j = 0; j < n; j++) {
      cout << ans[i][j] << " ";
    }
    cout << "\n";
  }
  return;
}
int main() {
  long long t = 1;
  while (t--) solve();
  return 0;
}