user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x7d000,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    $0x7d000,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movl   $0x0,-0x2c(%rbp)
  movl   $0x0,-0x28(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x1c(%rbp)
  jmp    L149
L148:
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x28(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x10(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L19
  addl   $0x1,-0x2c(%rbp)
  addl   $0x1,-0x28(%rbp)
L149:
  mov    -0x2c(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jge    L46
  mov    -0x28(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L148
  jmp    L46
L19:
  nop
L46:
  mov    -0x2c(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jne    L51
  mov    -0x20(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jne    L51
  mov    -0x30(%rbp),%eax
  cmp    %eax,-0x20(%rbp)
  jg     L51
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L34
L51:
  mov    -0x20(%rbp),%eax
  sub    -0x2c(%rbp),%eax
  mov    %eax,%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  sub    -0x2c(%rbp),%eax
  mov    %eax,-0x24(%rbp)
  jmp    L135
L68:
  mov    -0x2c(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x2c(%rbp)
  test   %eax,%eax
  jle    L60
  addl   $0x2,-0x24(%rbp)
  jmp    L135
L60:
  addl   $0x1,-0x24(%rbp)
L135:
  mov    -0x30(%rbp),%eax
  cmp    %eax,-0x24(%rbp)
  jl     L68
  mov    -0x30(%rbp),%eax
  cmp    %eax,-0x24(%rbp)
  jne    L97
  lea    D37(%rip),%rax
  jmp    L90
L97:
  lea    D30(%rip),%rax
L90:
  mov    %rax,%rdi
  call   <printf@plt>
L34:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L158
  call   <__stack_chk_fail@plt>
L158:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D2
type: string
name: parse_data
arguments: label: D29
type: string
name: parse_data
arguments: label: D8
type: string
name: parse_data
arguments: label: D37
type: string
name: parse_data
arguments: label: D30
type: string

tool: D2: %s

tool: D29: %d

tool: D8: Yes


tool: D37: Yes

tool: D30: No

assistant: ```c
int main() {
  char *s = (char *)malloc(512000 * sizeof(char));
  scanf("%s", s);
  char *t = (char *)malloc(512000 * sizeof(char));
  scanf("%s", t);
  int k;
  scanf("%d", &k);
  int i = 0;
  int j = 0;
  int lenT, lenS;
  lenT = strlen(t);
  lenS = strlen(s);
  while (i < lenS && j < lenT) {
    if (s[i] == t[j]) {
      i++;
      j++;
    } else {
      break;
    }
  }
  if (i == lenS && lenT == lenS && k >= lenT) {
    printf("Yes\n");
  } else {
    int p = lenT - i + lenS - i;
    while (p < k) {
      if (i-- > 0)
        p += 2;
      else
        p++;
    }
    printf(p == k ? "Yes" : "No");
  }
  return 0;
}
```