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[200][200];
int m,n;
void fun(int x,int y)
{
    for(int i=1;i<=m;i++)
        a[i][y]=0;
    for(int j=1;j<=n;j++)
        a[x][j]=0;
}
int main()
{
    int x[105],y[105];
    int b[200][200];
    int i,j;
    cin>>m>>n;
    for(i=1;i<=m;i++)
    {
        for(j=1;j<=n;j++)
            a[i][j]=1;
    }
    memset(x,0,sizeof(x));
    memset(y,0,sizeof(y));
    for(i=1;i<=m;i++)
    {
        for(j=1;j<=n;j++)
        {
            cin>>b[i][j];
            if(b[i][j]==0)
                fun(i,j);
        }
    }
    /*for(i=1;i<=m;i++)
    {
        for(j=1;j<=n;j++)
            cout<<a[i][j]<<" ";
        cout<<endl;
    }*/
    for(i=1;i<=m;i++)
    {
        for(j=1;j<=n;j++)
        {
            if(a[i][j]==1)
            {
                x[i]=1;
                y[j]=1;
            }
        }
    }
    /*cout<<x[1]<<" "<<x[2]<<endl;
    cout<<y[1]<<" "<<y[2]<<endl;*/
    int flag=0;
    for(i=1;i<=m;i++)
    {
        for(j=1;j<=n;j++)
        {
            if(x[i]==1||y[j]==1)
            {
                if(b[i][j]==1)
                    flag=1;
                else
                {
                    cout<<"NO"<<endl;
                    return 0;
                }
            }
            else
            {
                if(b[i][j]==1)
                {
                    cout<<"NO"<<endl;
                    return 0;
                }
                else
                    flag=1;
            }
        }
    }
    if(flag==1)
        cout<<"YES"<<endl;
    for(i=1;i<=m;i++)
    {
        for(j=1;j<n;j++)
            cout<<a[i][j]<<" ";
        cout<<a[i][n]<<endl;
    }
    return 0;
}