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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
T _abs(T n) {
  return (n < 0 ? -n : n);
}
template <class T>
T _max(T a, T b) {
  return (!(a < b) ? a : b);
}
template <class T>
T _min(T a, T b) {
  return (a < b ? a : b);
}
template <class T>
T sq(T n) {
  return n * n;
}
template <class T>
T gcd(T a, T b) {
  return (b != 0 ? gcd<T>(b, a % b) : a);
}
template <class T>
T lcm(T a, T b) {
  return (a / gcd<T>(a, b) * b);
}
template <class T>
T power(T N, T P) {
  if (P == 0) return 1;
  return (P == 1) ? N : N * power(N, P - 1);
}
template <class T>
double dist(T a, T b) {
  return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
template <class T>
string tostring(T n) {
  stringstream second;
  second << n;
  return second.str();
}
template <class T>
string itoa(T a) {
  if (!a) return "0";
  string ret;
  for (T i = a; i > 0; i = i / 10) ret.push_back((i % 10) + 48);
  reverse(ret.begin(), ret.end());
  return ret;
}
template <class T>
inline T mod(T n, T m) {
  return (n % m + m) % m;
}
int test, Case = 0;
int A[107][107], B[107][107], m, n;
vector<pair<int, int> > v;
void PUT0(int x, int y) {
  for (int i = 1; i <= n; i++) {
    A[x][i] = 0;
  }
  A[x][n + 1] = 0;
  for (int j = 1; j <= m; j++) {
    A[j][y] = 0;
  }
  A[m + 1][y] = 0;
}
bool check(int x, int y) {
  for (int i = 1; i <= n; i++) {
    if (A[x][i] == -1) return true;
  }
  for (int j = 1; j <= m; j++) {
    if (A[j][y] == -1) return true;
  }
  return false;
}
int main() {
  scanf("%d", &m);
  scanf("%d", &n);
  memset(A, -1, sizeof(A));
  for (int i = 1; i <= m; i++) {
    for (int j = 1; j <= n; j++) {
      scanf("%d", &B[i][j]);
      if (!B[i][j])
        PUT0(i, j);
      else
        v.push_back(make_pair(i, j));
    }
  }
  bool possible = true;
  for (long long i = 0; i < (int)v.size(); i++) {
    int x = v[i].first;
    int y = v[i].second;
    if (check(x, y))
      continue;
    else {
      possible = false;
      break;
    }
  }
  if (!possible)
    puts("NO");
  else {
    puts("YES");
    for (int i = 1; i <= m; i++) {
      for (int j = 1; j <= n; j++) {
        if (A[i][j] == -1)
          printf("%d", -1 * A[i][j]);
        else
          printf("%d", A[i][j]);
        if (j != n) printf(" ");
      }
      puts("");
    }
  }
  return 0;
}