// 確率的（mod 1000000007）解法
#include <iostream>
#include <stdio.h>
#include <stdint.h>
#include <cassert>
using namespace std;

using ll = long long;

#define TYPE_OF( VAR ) remove_const<remove_reference<decltype( VAR )>::type >::type
#define UNTIE ios_base::sync_with_stdio( false ); cin.tie( nullptr ) 
#define CEXPR( LL , BOUND , VALUE ) constexpr const LL BOUND = VALUE 
#define CIN( LL , A ) LL A; cin >> A
#define ASSERT( A , MIN , MAX ) assert( ( MIN ) <= A && A <= ( MAX ) )
#define CIN_ASSERT( A , MIN , MAX ) CIN( TYPE_OF( MAX ) , A ); ASSERT( A , MIN , MAX )
#define FOR( VAR , INITIAL , FINAL_PLUS_ONE ) for( TYPE_OF( FINAL_PLUS_ONE ) VAR = INITIAL ; VAR < FINAL_PLUS_ONE ; VAR ++ ) 
#define FOREQ( VAR , INITIAL , FINAL ) for( TYPE_OF( FINAL ) VAR = INITIAL ; VAR <= FINAL ; VAR ++ ) 
#define QUIT return 0 
#define RETURN( ANSWER ) cout << ( ANSWER ) << "\n"; QUIT 

#define POWER_MOD( ANSWER , ARGUMENT , EXPONENT , MODULO )		\
  ll ANSWER{ 1 };							\
  {									\
    ll ARGUMENT_FOR_SQUARE_FOR_POWER = ( MODULO + ( ( ARGUMENT ) % MODULO ) ) % MODULO; \
    TYPE_OF( EXPONENT ) EXPONENT_FOR_SQUARE_FOR_POWER = ( EXPONENT );	\
    while( EXPONENT_FOR_SQUARE_FOR_POWER != 0 ){			\
      if( EXPONENT_FOR_SQUARE_FOR_POWER % 2 == 1 ){			\
	ANSWER = ( ANSWER * ARGUMENT_FOR_SQUARE_FOR_POWER ) % MODULO;	\
      }									\
      ARGUMENT_FOR_SQUARE_FOR_POWER = ( ARGUMENT_FOR_SQUARE_FOR_POWER * ARGUMENT_FOR_SQUARE_FOR_POWER ) % MODULO; \
      EXPONENT_FOR_SQUARE_FOR_POWER /= 2;				\
    }									\
  }									\

inline CEXPR( int , bound_size , 15 );
// inline CEXPR( ll , P , 998244353 );
inline CEXPR( ll , P , 1000000007 );

int Rank( ll ( &A )[bound_size][bound_size] , const int& L , const int& M )
{
  int i_min = 0;
  int i_curr;
  int j_curr = 0;
  while( i_min < L && j_curr < M ){
    i_curr = i_min;
    while( i_curr < L ? A[i_curr][j_curr] == 0 : false ){
      i_curr++;
    }
    if( i_curr < L ){
      swap( A[i_min] , A[i_curr] );
      ll ( &A_i_min )[bound_size] = A[i_min];
      POWER_MOD( inv , A_i_min[j_curr] , P - 2 , P );
      FOR( j , j_curr , M ){
	( A_i_min[j] *= inv ) %= P;
      }
      FOR( i , i_min + 1 , L ){
	ll ( &A_i )[bound_size] = A[i];
	ll A_i_j_curr = A_i[j_curr];
	FOR( j , j_curr , M ){
	  ( A_i[j] -= A_i_j_curr * A_i_min[j] ) %= P;
	}
      }
      i_min++;
    }
    j_curr++;
  }
  return i_min;
}

int main()
{
  UNTIE;
  CIN_ASSERT( L , 1 , bound_size );
  CIN_ASSERT( M , 1 , bound_size / L );
  CIN_ASSERT( N , 1 , bound_size / M );
  CEXPR( ll , bound , 6 );
  ll A[bound_size][bound_size];
  FOR( i , 0 , L ){
    ll ( &Ai )[bound_size] = A[i];
    FOR( j , 0 , M ){
      CIN_ASSERT( Aij , - bound , bound );
      Ai[j] = Aij < 0 ? Aij += P : Aij;
    }
  }
  ll B[bound_size][bound_size];
  FOR( j , 0 , M ){
    ll ( &Bj )[bound_size] = B[j];
    FOR( k , 0 , N ){
      CIN_ASSERT( Bjk , - bound , bound );
      Bj[k] = Bjk < 0 ? Bjk += P : Bjk;
    }
  }
  ll sum = 0;
  FOR( i , 0 , L ){
    ll ( &Ai )[bound_size] = A[i];
    FOR( k , 0 , N ){
      FOR( j , 0 , M ){
	( sum += Ai[j] * B[j][k] ) %= P;
      }
      if( sum != 0 ){
	RETURN( "No" );
      }
    }
  }
  int rankA = Rank( A , L , M );
  int rankB = Rank( B , M , N );
  RETURN( rankB == M - rankA ? "Yes" : "No" );
}