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 JAVA):

Solution:
var size = readline().trim().split(' ');
var m = size[0], n = size[1];
var MatrixA = [], MatrixB = [];

for (var i = 0; i < m; i++) {
  MatrixB.push([]);
  MatrixB[i] = readline().trim().split(' ');
  for (var j = 0; j < n; j++) {
    MatrixB[i][j] = parseInt(MatrixB[i][j]);
  }
}
for (var i = 0; i < m; i++) {
  MatrixA.push([]);
  for (var j = 0; j < n; j++) {
    MatrixA[i][j] = 1;
  }
}


function setRowZero(row) {
  for (var i = 0; i < n; i++) {
    MatrixA[row][i] = 0;
  }
}
function setColumnZero(col) {
  for (var i = 0; i < m; i++) {
    MatrixA[i][col] = 0;
  }
}

for (var i = 0; i < m; i++) {
  for (var j = 0; j < n; j++) {
    if (MatrixB[i][j] === 0) {
      setRowZero(i);
      setColumnZero(j);
    }
  }
}

function check() {
  var isOk = true;
  for (var i = 0; i < m; i++) {
    for (var j = 0; j < n; j++) {
      var target = MatrixB[i][j];
      var source = 0;
      for (var k = 0; k < m; k++) {
        if (MatrixA[k][j] === 1) {
          source = 1;
        }
      }
      for (var k = 0; k < n; k++) {
        if (MatrixA[i][k] === 1) {
          source = 1;
        }
      }
      if (target !== source) {
        isOk = false;
      }
    }
  }
  if (isOk) {
    print('YES');
    for (var i = 0; i < m; i++) {
      print(MatrixA[i].join(' '));
    }
  } else {
    print('NO');
  }

}

check();