Problem: 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:

<image> where <image> 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:

<image>.

(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 m rows 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

Solution (in PYTHON3):

Solution:
import math,sys
from collections import defaultdict,deque
import bisect as bi
def yes():print('YES')
def no():print('NO')
def I():return (int(sys.stdin.readline()))
def In():return(map(int,sys.stdin.readline().split()))
def Sn():return sys.stdin.readline().strip()
def Pr(x): sys.stdout.write(str(x)+'\n')
#sys.setrecursionlimit(1500)
def dict(a):
    d={} 
    for x in a:
        if d.get(x,-1)!=-1:
            d[x]+=1
        else:
            d[x]=1
    return d

#  Find leftmost value >= x #
def find_gte(a, x):
    i = bi.bisect_left(a, x)
    if i != len(a):
        return i
    else:            
        return -1
def main():
    try:
        n,m=In()
        l=[list(In()) for i in range(n)]
        dp=[[0 for i in range(m)]for j in range(n)]
        ok=True
        for i in range(n):
            for j in range(m):
                if l[i][j]==1:
                    hor,ver=0,0
                    for k in range(m):
                        if l[i][k]==1:
                            hor+=1
                    for k in range(n):
                        if l[k][j]==1:
                            ver+=1
                    # if ver!=n and hor==m:
                    #     continue
                    # elif ver==n and hor!=m:
                    #     continue
                    # elif ver==n and hor==m:
                    #     dp[i][j]=1
                    # else:
                    #     ok=False
                    if ver!=n and hor!=m:
                        ok=False
                    elif ver==n and hor==m:
                        dp[i][j]=1
        check=[[0 for i in range(m)]for j in range(n)]    
        for i in range(n):
            for j in range(m):
                t1=0
                for k in range(m):
                    t1|=dp[i][k]
                for k in range(n):
                    t1|=dp[k][j]
                if t1!=l[i][j]:
                    ok=False
        if not ok:
            no()
        else:
            yes()
            for x in dp:
                print(*x)
    except:
        pass
        
M = 998244353
P = 1000000007
 
if __name__ == '__main__':
    # for _ in range(I()):main()
    for _ in range(1):main()
#End#

#        ******************* All The Best *******************   #