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

```asm
<countSort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  movslq %edx,%r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  lea    0x0(,%r13,4),%edi
  sub    $0x18,%rsp
  call   <malloc@plt>
  test   %rax,%rax
  jne    L133
L88:
  add    $0x18,%rsp
  mov    $0xffffffff,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L133:
  lea    0x0(,%r12,4),%r8d
  mov    %rax,%rbp
  mov    %r8d,%edi
  mov    %r8d,0xc(%rsp)
  call   <malloc@plt>
  mov    %rax,%r15
  test   %rax,%rax
  je     L109
  test   %r13d,%r13d
  mov    0xc(%rsp),%r8d
  mov    %r13,%r14
  jle    L122
  lea    0x0(,%r13,4),%rdx
  xor    %esi,%esi
  mov    %rbp,%rdi
  mov    %r8d,0xc(%rsp)
  call   <memset@plt>
  mov    0xc(%rsp),%r8d
L122:
  test   %r12d,%r12d
  jle    L45
  lea    -0x1(%r12),%edx
  mov    %rbx,%rax
  mov    %rdx,%r13
  lea    0x4(%rbx,%rdx,4),%rsi
  xchg   %ax,%ax
L54:
  movslq (%rax),%rdx
  add    $0x4,%rax
  addl   $0x1,0x0(%rbp,%rdx,4)
  cmp    %rax,%rsi
  jne    L54
  cmp    $0x1,%r14d
  jle    L126
L13:
  mov    0x0(%rbp),%esi
  lea    0x4(%rbp),%rax
  mov    $0x1,%edx
  nopl   0x0(%rax,%rax,1)
L62:
  add    (%rax),%esi
  add    $0x1,%edx
  add    $0x4,%rax
  mov    %esi,-0x4(%rax)
  cmp    %edx,%r14d
  jg     L62
  test   %r13d,%r13d
  js     L86
L126:
  movslq %r13d,%rax
  nopl   0x0(%rax)
L55:
  movslq (%rbx,%rax,4),%rdx
  sub    $0x1,%rax
  mov    %rdx,%rcx
  lea    0x0(%rbp,%rdx,4),%rdx
  movslq (%rdx),%rsi
  mov    %ecx,-0x4(%r15,%rsi,4)
  subl   $0x1,(%rdx)
  test   %eax,%eax
  jns    L55
L86:
  mov    %r8d,%edx
  mov    %r15,%rsi
  mov    %rbx,%rdi
  call   <memcpy@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  mov    %r15,%rdi
  call   <free@plt>
  xor    %eax,%eax
  lea    D27(%rip),%rdi
  call   <printf@plt>
  test   %r12d,%r12d
  jle    L98
  mov    %r13d,%r13d
  lea    D6(%rip),%rbp
  lea    0x4(%rbx,%r13,4),%r12
  nopw   0x0(%rax,%rax,1)
L70:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L70
L98:
  add    $0x18,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    10b0 <putchar@plt>
  nopl   (%rax)
L109:
  mov    %rbp,%rdi
  call   <free@plt>
  jmp    L88
  nopl   (%rax)
L45:
  lea    -0x1(%r12),%r13d
  cmp    $0x1,%r14d
  jg     L13
  mov    %r8d,%edx
  mov    %r15,%rsi
  mov    %rbx,%rdi
  call   <memcpy@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  mov    %r15,%rdi
  call   <free@plt>
  lea    D27(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L98
```

tool_calls: name: parse_data
arguments: tag: D27
type: string
name: parse_data
arguments: tag: D6
type: string

tool: D27: Count sort 

tool: D6: %3d

assistant: ```c
int countSort(int *data, int size, int k) {
  int *counts, *temp;
  int i, j;
  if ((counts = (int *)malloc(k * sizeof(int))) == NULL)
    return -1;
  if ((temp = (int *)malloc(size * sizeof(int))) == NULL) {
    free(counts);
    return -1;
  }
  for (i = 0; i < k; i++)
    counts[i] = 0;
  for (j = 0; j < size; j++)
    counts[data[j]] = counts[data[j]] + 1;
  for (i = 1; i < k; i++)
    counts[i] = counts[i] + counts[i - 1];
  for (j = size - 1; j >= 0; j--) {
    temp[counts[data[j]] - 1] = data[j];
    counts[data[j]] = counts[data[j]] - 1;
  }
  memcpy(data, temp, size * sizeof(int));
  free(counts);
  free(temp);
  printf("Count sort ");
  for (i = 0; i < size; i++) {
    printf("%3d", data[i]);
  }
  printf("\n");
}
```