user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x8e0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  jmp    L103
L91:
  lea    -0x8c0(%rbp),%rax
  mov    $0x68,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  lea    -0x850(%rbp),%rax
  mov    $0x68,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  lea    -0x7e0(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x8cc(%rbp)
  lea    -0x3f0(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x8c8(%rbp)
  movl   $0x0,-0x8d4(%rbp)
  jmp    L35
L20:
  mov    -0x8d4(%rbp),%eax
  cltq
  movzbl -0x7e0(%rbp,%rax,1),%eax
  mov    %al,-0x8d5(%rbp)
  movsbl -0x8d5(%rbp),%eax
  sub    $0x61,%eax
  movslq %eax,%rdx
  mov    -0x8c0(%rbp,%rdx,4),%edx
  add    $0x1,%edx
  cltq
  mov    %edx,-0x8c0(%rbp,%rax,4)
  addl   $0x1,-0x8d4(%rbp)
L35:
  mov    -0x8d4(%rbp),%eax
  cmp    -0x8cc(%rbp),%eax
  jl     L20
  movl   $0x0,-0x8d4(%rbp)
  jmp    L26
L76:
  mov    -0x8d4(%rbp),%eax
  cltq
  movzbl -0x3f0(%rbp,%rax,1),%eax
  mov    %al,-0x8d5(%rbp)
  movsbl -0x8d5(%rbp),%eax
  sub    $0x61,%eax
  movslq %eax,%rdx
  mov    -0x850(%rbp,%rdx,4),%edx
  add    $0x1,%edx
  cltq
  mov    %edx,-0x850(%rbp,%rax,4)
  addl   $0x1,-0x8d4(%rbp)
L26:
  mov    -0x8d4(%rbp),%eax
  cmp    -0x8c8(%rbp),%eax
  jl     L76
  movl   $0x0,-0x8d4(%rbp)
  jmp    L123
L92:
  mov    -0x8d4(%rbp),%eax
  cltq
  mov    -0x8c0(%rbp,%rax,4),%eax
  test   %eax,%eax
  je     L31
  mov    -0x8d4(%rbp),%eax
  cltq
  mov    -0x850(%rbp,%rax,4),%eax
  test   %eax,%eax
  je     L31
  mov    -0x8d4(%rbp),%eax
  cltq
  mov    -0x8c0(%rbp,%rax,4),%edx
  mov    -0x8d4(%rbp),%eax
  cltq
  mov    -0x850(%rbp,%rax,4),%eax
  cmp    %eax,%edx
  cmovle %edx,%eax
  mov    %eax,-0x8c4(%rbp)
  movl   $0x0,-0x8d0(%rbp)
  jmp    L109
L46:
  mov    -0x8d4(%rbp),%eax
  add    $0x61,%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
  addl   $0x1,-0x8d0(%rbp)
L109:
  mov    -0x8d0(%rbp),%eax
  cmp    -0x8c4(%rbp),%eax
  jl     L46
L31:
  addl   $0x1,-0x8d4(%rbp)
L123:
  cmpl   $0x19,-0x8d4(%rbp)
  jle    L92
  mov    $0xa,%edi
  call   <putchar@plt>
L103:
  lea    -0x7e0(%rbp),%rax
  mov    %rax,%rdi
  call   <gets@plt>
  test   %rax,%rax
  je     L116
  lea    -0x3f0(%rbp),%rax
  mov    %rax,%rdi
  call   <gets@plt>
  test   %rax,%rax
  jne    L91
L116:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L70
  call   <__stack_chk_fail@plt>
L70:
  leave
  ret
```

assistant: ```c
int main() {
  char arg1[1000], arg2[1000];
  int alpha_stat_arg1[26], alpha_stat_arg2[26];
  int len_arg1, len_arg2;
  int i, j, count;
  char ch;
  while (gets(arg1) && gets(arg2)) {
    memset(alpha_stat_arg1, 0x00, sizeof(int) * 26);
    memset(alpha_stat_arg2, 0x00, sizeof(int) * 26);
    len_arg1 = strlen(arg1);
    len_arg2 = strlen(arg2);
    for (i = 0; i < len_arg1; i++) {
      ch = arg1[i];
      alpha_stat_arg1[ch - 'a']++;
    }
    for (i = 0; i < len_arg2; i++) {
      ch = arg2[i];
      alpha_stat_arg2[ch - 'a']++;
    }
    for (i = 0; i < 26; i++) {
      if (alpha_stat_arg1[i] && alpha_stat_arg2[i]) {
        count = ((alpha_stat_arg1[i]) > (alpha_stat_arg2[i])) ? (alpha_stat_arg2[i]) : (alpha_stat_arg1[i]);
        for (j = 0; j < count; j++) {
          putchar(i + 'a');
        }
      }
    }
    putchar('\n');
  }
  return 0;
}
```