OR in Matrix

``` A B 都是 n*m 的矩阵，已知 B 矩阵是由 A 矩阵以某种规则生成， B [i][j]是由 A 矩阵的第 i 行的所有元素和第 j 列的所有元素 或运算（or）得到 ```  ``` 给定 B 矩阵，求是否存在一个矩阵 A 1 <= n,m <= 100 ``` 感谢@凉凉 提供的翻译

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:  where  is equal to $ 1 $ if some $ a_{i}=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 $ A_{ij} $ . All elements of $ A $ are either 0 or 1. From matrix $ A $ , Nam creates another matrix $ B $ of the same size using formula: . ( $ B_{ij} $ 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.

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 $ ).

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.

2 2
1 0
0 0

NO

2 3
1 1 1
1 1 1

YES
1 1 1
1 1 1

2 3
0 1 0
1 1 1

YES
0 0 0
0 1 0