codeforces 486 B. OR in Matrix

B. OR in Matrix

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

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

或运算的运用，“全假出假”，如果b数组元素有0，那么那一行和那一列全为0，标记一下，然后求出数组a的每个元素，再验证是否能得到数组b

代码：

#include 
#include 
#include 
using namespace std;

int main()
{
    int n,m,a[110][110],b[110][110];
    int row[110],col[110];
    scanf("%d%d",&n,&m);
    memset(row,0,sizeof(row));
    memset(col,0,sizeof(col));
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            scanf("%d",&b[i][j]);
            if(b[i][j]==0)  //标记全为0 的行列
            {
                row[i]=1;
                col[j]=1;
            }
        }
    }
    for(int i=1;i<=n;i++)   //求数组a
    {
        for(int j=1;j<=m;j++)
        {
            if(row[i]||col[j])
                a[i][j]=0;
            else
                a[i][j]=1;
        }
    }
    memset(row,0,sizeof(row));
    memset(col,0,sizeof(col));
    //求出行列的或运算值
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            row[i]|=a[i][j];
        }
    }
    for(int j=1;j<=m;j++)
    {
        for(int i=1;i<=n;i++)
            col[j]|=a[i][j];
    }
    //验证是否能得到b数组
    int flag=0;
    for(int i=1;i<=n;i++)
    {
        if(flag)
            break;
        for(int j=1;j<=m;j++)
        {
            if(flag)
                break;
            if(b[i][j]!=(row[i]|col[j]))
                flag=1;
        }
    }
    //输出
    if(flag)
        printf("NO\n");
    else
    {
        printf("YES\n");
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                if(j==1) printf("%d",a[i][j]);
                else printf(" %d",a[i][j]);
            }
            printf("\n");
        }
    }
    return 0;
}