B. OR in Matrix

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 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:

.

(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 mrows consisting of n integers representing matrix A that can produce given matrix B. If there are several solutions print any one.

Sample test(s)

input

2 21 00 0

output

NO

input

2 31 1 11 1 1

output

YES1 1 11 1 1

input

2 30 1 01 1 1

output

YES0 0 00 1 0

这题刚开始以为要用搜索，看了别人的题解才知道不用的，因为只要把0所在的行和列都变成0，其他的都变为1就得到了原来的矩阵A,（注意：其实由B得到的矩阵不止A这一个，但是这样得到的A是所有的到矩阵的最优解，因为得到1的数是最多的。

#include<stdio.h>#include<string.h>int a[200][200],b[200][200],c[200][200];int main(){int n,m,i,j,h,t,flag,flag1;while(scanf("%d%d",&n,&m)!=EOF){memset(a,0,sizeof(a));memset(b,0,sizeof(b));memset(c,0,sizeof(c));for(i=1;i<=n;i++){for(j=1;j<=m;j++){scanf("%d",&a[i][j]);b[i][j]=1;}}for(i=1;i<=n;i++){for(j=1;j<=m;j++){if(a[i][j]==0){for(h=1;h<=m;h++){b[i][h]=0;}for(t=1;t<=n;t++){b[t][j]=0;}}}}for(i=1;i<=n;i++){for(j=1;j<=m;j++){flag=0;for(h=1;h<=m;h++){if(b[i][h]==1){flag=1;break;}}for(t=1;t<=n;t++){if(b[t][j]==1){flag=1;break;}}if(flag==1){c[i][j]=1;continue;}c[i][j]=0;}}flag1=1;for(i=1;i<=n;i++){for(j=1;j<=m;j++){if(c[i][j]!=a[i][j]){flag1=0;break;}}if(flag1==0)break;}if(flag1==0){printf("NO\n");continue;}printf("YES\n");for(i=1;i<=n;i++){for(j=1;j<=m;j++){if(j!=m)printf("%d ",b[i][j]);else printf("%d\n",b[i][j]);}}}return 0;}