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;
template <typename T>
using vec = std::vector<T>;
const int INF = 1 << 30;
const long long int INF_ = 1LL << 58;
const double EPS = 1e-9;
const int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &vec) {
  os << "[";
  for (const auto &v : vec) {
    os << v << ",";
  }
  os << "]";
  return os;
}
int m, n;
void Solve() {
  cin >> m >> n;
  bool f = true;
  vector<vector<int>> v(m, vector<int>(n)), ans(m, vector<int>(n, -1));
  for (int y = (0); y < (m); ++y) {
    for (int x = (0); x < (n); ++x) {
      cin >> v[y][x];
    }
  }
  for (int y1 = (0); y1 < (m); ++y1)
    for (int x1 = (0); x1 < (n); ++x1) {
      if (v[y1][x1] == 1) continue;
      for (int y2 = (0); y2 < (m); ++y2) {
        ans[y2][x1] = 0;
      }
      for (int x2 = (0); x2 < (n); ++x2) {
        ans[y1][x2] = 0;
      }
    }
  for (int y1 = (0); y1 < (m); ++y1)
    for (int x1 = (0); x1 < (n); ++x1) {
      if (v[y1][x1] == 0) continue;
      bool ff = false;
      for (int y2 = (0); y2 < (m); ++y2) {
        if (ans[y2][x1] != 0) {
          ans[y2][x1] = 1;
          ff = true;
        }
      }
      for (int x2 = (0); x2 < (n); ++x2) {
        if (ans[y1][x2] != 0) {
          ans[y1][x2] = 1;
          ff = true;
        }
      }
      if (!ff) {
        ;
        ;
        f = false;
        break;
      }
    }
  if (f) {
    cout << "YES" << endl;
    for (int y = (0); y < (m); ++y)
      for (int x = (0); x < (n); ++x)
        cout << ans[y][x] << (x == n - 1 ? '\n' : ' ');
  } else
    cout << "NO" << endl;
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  Solve();
  return 0;
}