CodeForces 486B

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.

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.

Sample Input

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
第一次看题目把a，b两个数组看反了，我觉得这道题还是挺好写的，反这写就好了，只要判断a数组当行列上有1的时候对应的b数组是否为1，首先要把a数组复制出来
b数组中为0是，a数组对应的行列都要为0；

 #include<iostream>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 using namespace std;
 int a[][];
 int m,n;
 void fun(int x,int y)
 {
     for(int i=;i<=m;i++)
         a[i][y]=;
     for(int j=;j<=n;j++)
         a[x][j]=;
 }
 int main()
 {
     int x[],y[];
     int b[][];
     int i,j;
     cin>>m>>n;
     for(i=;i<=m;i++)
     {
         for(j=;j<=n;j++)
             a[i][j]=;
     }
     memset(x,,sizeof(x));
     memset(y,,sizeof(y));
     for(i=;i<=m;i++)
     {
         for(j=;j<=n;j++)
         {
             cin>>b[i][j];
             if(b[i][j]==)
                 fun(i,j);
         }
     }
     /*for(i=1;i<=m;i++)
     {
         for(j=1;j<=n;j++)
             cout<<a[i][j]<<" ";
         cout<<endl;
     }*/
     for(i=;i<=m;i++)
     {
         for(j=;j<=n;j++)
         {
             if(a[i][j]==)
             {
                 x[i]=;
                 y[j]=;
             }
         }
     }
     /*cout<<x[1]<<" "<<x[2]<<endl;
     cout<<y[1]<<" "<<y[2]<<endl;*/
     int flag=;
     for(i=;i<=m;i++)
     {
         for(j=;j<=n;j++)
         {
             if(x[i]==||y[j]==)
             {
                 if(b[i][j]==)
                     flag=;
                 else
                 {
                     cout<<"NO"<<endl;
                     return ;
                 }
             }
             else
             {
                 if(b[i][j]==)
                 {
                     cout<<"NO"<<endl;
                     return ;
                 }
                 else
                     flag=;
             }
         }
     }
     if(flag==)
         cout<<"YES"<<endl;
     for(i=;i<=m;i++)
     {
         for(j=;j<n;j++)
             cout<<a[i][j]<<" ";
         cout<<a[i][n]<<endl;
     }
     return ;
 }