CF486B.OR in Matrix

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.

我们定义逻辑 OR（或运算）为作用于两个逻辑值（即属于集合 $\{0, 1\}$ 的值）上的运算：当且仅当两个逻辑值中至少有一个为 $1$ 时，结果为 $1$；否则结果为 $0$。类似地，我们可以将逻辑 OR 扩展到三个或更多逻辑值：

其中 在某个 $a_i = 1$ 时等于 $1$，否则等于 $0$。

Nam 拥有一个由 $m$ 行和 $n$ 列组成的矩阵 $A$。行编号从 $1$ 到 $m$，列编号从 $1$ 到 $n$。位于第 $i$ 行（$1 ≤ i ≤ m$）和第 $j$ 列（$1 ≤ j ≤ n$）的元素记为 $A_{ij}$。矩阵 $A$ 中所有元素均为 $0$ 或 $1$。Nam 根据矩阵 $A$ 构造另一个同尺寸的矩阵 $B$，其定义如下：

。

（即 $B_{ij}$ 是矩阵 $A$ 中第 $i$ 行与第 $j$ 列所有元素的逻辑 OR）

Nam 给出矩阵 $B$，并挑战你反推出矩阵 $A$。尽管 Nam 很聪明，但由于矩阵 $A$ 的尺寸可能很大，他在计算矩阵 $B$ 时仍有可能出错。

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).

第一行包含两个整数 $m$ 和 $n$（$1 \leq m, n \leq 100$），分别表示矩阵的行数和列数。

接下来的 $m$ 行，每行包含 $n$ 个用空格分隔的整数，描述矩阵 $B$ 的各行（$B$ 的每个元素为 $0$ 或 $1$）。

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.

第一行，如果 Nam 在计算矩阵 $B$ 时出错，则输出 "NO"；否则输出 "YES"。若第一行输出为 "YES"，则还需输出 $m$ 行，每行包含 $n$ 个整数，表示一个能生成给定矩阵 $B$ 的矩阵 $A$。若存在多个解，输出任意一个即可。