user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x24(%rbp)
  movl   $0x0,-0x20(%rbp)
  movl   $0x0,-0x1c(%rbp)
  movl   $0x0,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x24(%rbp),%rax
  mov    %rax,%rsi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L56
L121:
  mov    -0x24(%rbp),%eax
  cmp    $0x63,%eax
  jle    L90
  mov    -0x24(%rbp),%eax
  sub    $0x64,%eax
  mov    %eax,-0x24(%rbp)
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L56
L90:
  mov    -0x24(%rbp),%eax
  cmp    $0x31,%eax
  jle    L99
  mov    -0x24(%rbp),%eax
  sub    $0x32,%eax
  mov    %eax,-0x24(%rbp)
  mov    -0x10(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x10(%rbp)
  jmp    L56
L99:
  mov    -0x24(%rbp),%eax
  cmp    $0x13,%eax
  jle    L16
  mov    -0x24(%rbp),%eax
  sub    $0x14,%eax
  mov    %eax,-0x24(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x14(%rbp)
  jmp    L56
L16:
  mov    -0x24(%rbp),%eax
  cmp    $0x9,%eax
  jle    L81
  mov    -0x24(%rbp),%eax
  sub    $0xa,%eax
  mov    %eax,-0x24(%rbp)
  mov    -0x18(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x18(%rbp)
  jmp    L56
L81:
  mov    -0x24(%rbp),%eax
  cmp    $0x4,%eax
  jle    L46
  mov    -0x24(%rbp),%eax
  sub    $0x5,%eax
  mov    %eax,-0x24(%rbp)
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x1c(%rbp)
  jmp    L56
L46:
  mov    -0x24(%rbp),%eax
  test   %eax,%eax
  jle    L56
  mov    -0x24(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x24(%rbp)
  mov    -0x20(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x20(%rbp)
L56:
  mov    -0x24(%rbp),%eax
  test   %eax,%eax
  jg     L121
  mov    -0x20(%rbp),%esi
  mov    -0x1c(%rbp),%r8d
  mov    -0x18(%rbp),%edi
  mov    -0x14(%rbp),%ecx
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  sub    $0x8,%rsp
  push   %rsi
  mov    %r8d,%r9d
  mov    %edi,%r8d
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x10,%rsp
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L75
  call   <__stack_chk_fail@plt>
L75:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D26
data_type: string
name: unpack_data
arguments: data_tag: D17
data_type: string
name: unpack_data
arguments: data_tag: D6
data_type: string

tool: D26: Please enter the amount of money you wish to withdraw: 

tool: D17: %d

tool: D6: You recieved %d hundred(s)
 You recieved %d fifty(s)
 You recieved %d twenty(s)
 You recieved %d ten(s)
 You recieved %d five(s)
 You recieved %d one(s)


assistant: ```c
int main() {
  int numChange = 0;
  int billsChange[] = {0, 0, 0, 0, 0, 0};
  printf("Please enter the amount of money you wish to withdraw: ");
  scanf("%d", &numChange);
  while (numChange > 0) {
    if (numChange >= 100) {
      numChange -= 100;
      billsChange[5]++;
    } else if (numChange >= 50) {
      numChange -= 50;
      billsChange[4]++;
    } else if (numChange >= 20) {
      numChange -= 20;
      billsChange[3]++;
    } else if (numChange >= 10) {
      numChange -= 10;
      billsChange[2]++;
    } else if (numChange >= 5) {
      numChange -= 5;
      billsChange[1]++;
    } else if (numChange >= 1) {
      numChange -= 1;
      billsChange[0]++;
    }
  }
  printf("You recieved %d hundred(s)\n You recieved %d fifty(s)\n You recieved %d twenty(s)\n You recieved %d ten(s)\n You recieved %d five(s)\n You recieved %d one(s)\n", billsChange[5], billsChange[4], billsChange[3], billsChange[2], billsChange[1], billsChange[0]);
  return 0;
}
```