// Problem         : Hammering (NAIPC 2019)
// Author          : Darcy Best
// Expected Result : AC
// Complexity      : O(rc)

//  ========= INITAL CHECK:
// - If the start and final configs are the same, then answer is YES.

//  ========= MARK HITTABLE ROWS:
// - In the fianl config, if 2 or more are down in any given row/col,
//   then that row is not hittable.
// - If there is a peg that needs X --> X and *one of* its row/col is not hittable,
//   then the other one isn't hittable either.
// - If there is a peg that needs O --> X, then *both* row and column
//   must be hittable.
// - If there is a peg that needs X --> O, then *at least one* of its row
//   or column must be hittable.

//  ========= FINAL CHECK:
// - In the region of hittable cells, if there are no X's in the final configuration,
//   then answer is NO.
// - In the region of hittable cells, if there are no O's in the inital configuration,
//   then answer is NO.
//
//  ========= DO THE HITTING:
// - The answer at this point is YES. Here is how:
// (1) Hit every cell in the hittable region so that there is only one peg up (one that
//     needs to be an X).
// (2) Hit each other X in the final configuration.

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;

bool chk(){
  int r,c; cin >> r >> c;
  
  vector<string> A(r), B(r);
  
  for(auto& x : A) cin >> x;
  for(auto& x : B) cin >> x;
  
  if(A == B) return true;

  vector<bool> hittable_row(r,true), hittable_col(c,true);

  for(int i=0;i<r;i++){
    int cnt = 0;
    for(int j=0;j<c;j++)
      if(B[i][j] == 'X')
        cnt++;
    if(cnt > 1) hittable_row[i] = false;
  }

  for(int j=0;j<c;j++){
    int cnt = 0;
    for(int i=0;i<r;i++)
      if(B[i][j] == 'X')
        cnt++;
    if(cnt > 1) hittable_col[j] = false;
  }

  for(int i=0;i<r;i++)
    for(int j=0;j<c;j++)
      if(A[i][j] == 'X' && B[i][j] == 'X'){
        if(!hittable_row[i]) hittable_col[j] = false;
        if(!hittable_col[j]) hittable_row[i] = false;
      }

  for(int i=0;i<r;i++)
    for(int j=0;j<c;j++)
      if(A[i][j] == 'O' && B[i][j] == 'X')
        if(!hittable_row[i] || !hittable_col[j])
          return false;

  for(int i=0;i<r;i++)
    for(int j=0;j<c;j++)
      if(A[i][j] == 'X' && B[i][j] == 'O')
        if(!hittable_row[i] && !hittable_col[j])
          return false;

  int AO_cnt = 0, BX_cnt = 0;
  for(int i=0;i<r;i++)
    for(int j=0;j<c;j++)
      if(hittable_row[i] && hittable_col[j]){
        if(A[i][j] == 'O') AO_cnt++;
        if(B[i][j] == 'X') BX_cnt++;
      }

  return AO_cnt > 0 && BX_cnt > 0;
}

int main(){
  cout << chk() << endl;
}