/*
# 问题分析
题目要求按照步骤进行计算, 并输出计算结果。
这种明确给出计算步骤的题目妥妥的"模拟题", 明确题设要求的计算步骤后, 翻译成代码即可。
 
# 加密步骤
以"1234567"和"368782971"为例, 加密过程为:
A        0 0 1 2 3 4 5 6 7 
B        3 6 8 7 8 2 9 7 1
Result   3 6 9 5 Q 8 1 1 8
1. 先将A和B通过将较小的数高位补0使得两个数可以对齐
2. 然后从开始按照奇数位偶数位规则计算
 
# 完整描述步骤
1. 获取输入: A和B
2. 补齐A和B的长度
3. 从最后一位开始, 按照规则计算
 
# 伪代码描述
1. get input: A, B
2. if length(A) < length(B):
    - fill 0 for A
   else if length(A) > length(B):
    - fill 0 for B
3. for each odd position:
    - result[odd] = (A[odd] + B[odd]) % 13
    - if result[odd] > 9:
        - map result[odd]
4. for each even position:
    - result[even] = B[even] - A[even]
    - if result[event] < 0:
        - result[even] += 10
*/
 
 
# include<stdio.h>
 
int main() {
  char A[101];
  char B[101];
  char result[101];
  scanf("%s %s", &A, &B);
  char signals[] = {'0', '1', '2', '3', '4', '5', '6',
                    '7', '8', '9', 'J', 'Q', 'K'};
 
  int A_size, B_size, max_size;
  for (A_size = 0; A[A_size]; A_size++)
    ;
  for (B_size = 0; B[B_size]; B_size++)
    ;
 
  if (A_size > B_size) {
    max_size = A_size;
    int offset = A_size - B_size;
    for (int i = B_size - 1; i >= 0; i--) {
      B[i + offset] = B[i];
    }
    for (int i = 0; i < offset; i++) {
      B[i] = '0';
    }
    B[A_size] = 0;
  } else if (A_size < B_size) {
    max_size = B_size;
    int offset = B_size - A_size;
    for (int i = A_size - 1; i >= 0; i--) {
      A[i + offset] = A[i];
    }
    for (int i = 0; i < offset; i++) {
      A[i] = '0';
    }
    A[B_size] = 0;
  }
 
  for (int i = max_size - 1; i >= 0; i = i - 2) {
    result[i] = signals[((A[i] - '0') + (B[i] - '0')) % 13];
  }
 
  int difference;
  for (int i = max_size - 2; i >= 0; i = i - 2) {
    difference = (B[i] - '0') - (A[i] - '0');
    if (difference < 0) {
      difference += 10;
    }
    result[i] = signals[difference];
  }
  result[max_size] = 0;
  printf("%s\n", result);
  return 0;
}