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;
const int MOD = 1e9 + 7;
template <class F, class S>
ostream &operator<<(ostream &out, const pair<F, S> &p) {
  return out << '(' << p.first << ", " << p.second << ')';
}
template <class T>
auto operator<<(ostream &out, T &&x) -> decltype(x.begin(), out) {
  out << '{';
  for (auto i = x.begin(); i != x.end(); i++)
    out << *i << (i == prev(x.end()) ? "" : ", ");
  return out << '}';
}
void dmp() {}
template <class T, class... Args>
void dmp(T &&x, Args... args) {
  cerr << x << ";  ";
  dmp(args...);
}
int r, c;
bool row[101], col[101], a[101][101], ans[101][101];
void solve() {
  cin >> r >> c;
  bool ok = 1;
  for (int i = 1; i <= r; i++) {
    for (int j = 1; j <= c; j++) {
      cin >> a[i][j];
      ans[i][j] = 1;
    }
  }
  for (int i = 1; i <= r; i++) {
    for (int j = 1; j <= c; j++) {
      if (!a[i][j]) {
        for (int k = 1; k <= c; k++) ans[i][k] = 0;
        for (int k = 1; k <= r; k++) ans[k][j] = 0;
      }
    }
  }
  bool f;
  for (int i = 1; i <= r; i++) {
    for (int j = 1; j <= c; j++) {
      f = 0;
      for (int k = 1; k <= c; k++) f |= ans[i][k];
      for (int k = 1; k <= r; k++) f |= ans[k][j];
      if (a[i][j] != f) {
        ok = 0;
        break;
      }
    }
  }
  if (ok) {
    cout << "YES" << '\n';
    for (int i = 1; i <= r; i++) {
      for (int j = 1; j <= c; j++) {
        cout << ans[i][j] << " \n"[j == c];
      }
    }
  } else
    cout << "NO" << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t = 1;
  while (t--) solve();
}