/* Sample solution for NWERC'06: Sudoku
 * Author: Per Austrin
 * Algorithm: 
 * Fix the position of one 3x3 block at a time, then try all
 * permutations within that block and proceed to the next 3x3-block.
 * Keep a partial remapping of the digits as we go, and break whenever
 * inconsistencies are found.
 */
#include <cstdio>
#include <algorithm>

using namespace std;

int b1[9][9], b2[9][9];
int brperm[3], bcperm[3];
int rperm[3][3], cperm[3][3];

bool tryperm(int blockrow, int blockcol, int *digmap) {
  if (blockcol == 3) ++blockrow, blockcol = 0;
  if (blockrow == 3) return true;

  for (int i = 0; i < 3; ++i)
    if (blockcol == 0 && brperm[i] == -1 || 
	blockcol > 0 && brperm[i] == blockrow)
      for (int j = 0; j < 3; ++j)
	if (blockrow == 0 && bcperm[j] == -1 ||
	    blockrow > 0 && bcperm[j] == blockcol) {
	  int rp[3], cp[3];
	  brperm[i] = blockrow;
	  bcperm[j] = blockcol;

	  for (int k = 0; k < 3; ++k) rp[k] = cp[k] = k;
	  // Check if any of these permutations have already been fixed.
	  if (blockrow > 0) memcpy(cp, cperm[blockcol], sizeof(cp));
	  if (blockcol > 0) memcpy(rp, rperm[blockrow], sizeof(rp));

	  do {
	    do {
	      // Check that row and column permutations for this block
	      // are consistent with the current remapping of the digits.
	      bool inconsistent = false;
	      int ndigmap[10];
	      memcpy(ndigmap, digmap, sizeof(int)*10);
	      for (int r = 0; !inconsistent && r < 3; ++r)
		for (int c = 0; !inconsistent && c < 3; ++c) {
		  int v1 = b1[3*blockrow + r][3*blockcol + c];	
		  int v2 = b2[3*i + rp[r]][3*j + cp[c]];
		  if (v2) {
		    if (ndigmap[v2] && ndigmap[v2] != v1) inconsistent = true;
		    ndigmap[v2] = v1;
		  }
		}
	      memcpy(cperm[blockcol], cp, sizeof(cp));
	      memcpy(rperm[blockrow], rp, sizeof(rp));
	      if (!inconsistent && 
		  tryperm(blockrow, blockcol+1, ndigmap))
		return true;
	    } while (blockrow == 0 && next_permutation(cp, cp+3));
	  } while (blockcol == 0 && next_permutation(rp, rp+3));

	  // Restore block permutations
	  if (blockcol == 0) brperm[i] = -1;
	  if (blockrow == 0) bcperm[j] = -1;
	}
  return false;
}

int main(void) {
  int T, i;
  scanf("%d", &T);
  for (; T--; ) {
    for (i = 0; i < 81; ++i) scanf(" %1d", b1[i/9]+i%9);
    for (i = 0; i < 81; ++i) scanf(" %1d", b2[i/9]+i%9);
    // Only need to check rotation by 90 degrees since additional
    // rotation by 180 degrees can be achieved by the permutations.
    for (i = 0; i < 2; ++i) {
      int digmap[10];
      int nb2[9][9];
      for (int r = 0; r < 9; ++r)
	for (int c = 0; c < 9; ++c)
	  nb2[r][c] = b2[c][8-r];
      memcpy(b2, nb2, sizeof(b2));
      memset(brperm, -1, sizeof(brperm));
      memset(bcperm, -1, sizeof(bcperm));
      memset(digmap, 0, sizeof(digmap));
      if (tryperm(0, 0, digmap)) {
	printf("Yes\n");
	break;
      }
    }
    if (i == 2) printf("No\n");
  }
  return 0;
}