OR in Matrix

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

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

/*题意：定义一个异或，a1^a2...^an如果有一个ai=1那么值为1，否则为零，给出你一个矩阵B,是由矩阵A得来的，Bij等于A的i行元素    异或j列元素。给出你矩阵B问你是否有这样的矩阵A初步思路：将矩阵初始化为1，然后先按照矩阵B中有零的元素，将对应A矩阵中的元素设置成零，然后在反过来验证B矩阵*/#include 
         
          using namespace std;int n,m;int a[110][110];int b[110][110];int main(){    // freopen("in.txt","r",stdin);    scanf("%d%d",&n,&m);    for(int i=1;i<=n;i++){        for(int j=1;j<=m;j++){            scanf("%d",&b[i][j]);            a[i][j]=1;        }    }    //按照B矩阵进行置0    for(int i=1;i<=n;i++){        for(int j=1;j<=m;j++){            if(b[i][j]==0){                for(int k=1;k<=m;k++)                    a[i][k]=0;                for(int k=1;k<=n;k++)                    a[k][j]=0;            }        }    }    //验证然后按照1的位置验证B矩阵    bool f=false;    for(int i=1;i<=n;i++){        for(int j=1;j<=m;j++){            if(b[i][j]==1){                bool flag=false;                for(int k=1;k<=n;k++){                    if(a[i][k]==1){                        flag=true;                        break;                    }                }                if(flag==false)                for(int k=1;k<=m;k++){                    if(a[k][j]==1){                        flag=true;                        break;                    }                }                if(flag==false){                    f=true;                    break;                }            }        }    }    if(f){        puts("NO");    }else{        puts("YES");        for(int i=1;i<=n;i++){            for(int j=1;j<=m;j++){                printf(j==1?"%d":" %d",a[i][j]);            }            printf("\n");        }    }    return 0;}