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

```asm
<ms>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  mov    %rsp,-0x48(%rbp)
  cmp    $0x1,%esi
  je     L154
  mov    %esi,%ecx
  mov    %esi,%r15d
  mov    %rdi,%r13
  shr    $0x1f,%ecx
  add    %esi,%ecx
  sar    %ecx
  mov    %ecx,%r14d
  mov    %ecx,%esi
  call   <ms>
  movslq %r14d,%rax
  mov    %r15d,%esi
  sub    %r14d,%esi
  lea    0x0(%r13,%rax,4),%rdi
  call   <ms>
  movslq %r15d,%r10
  mov    %rsp,%rsi
  lea    0xf(,%r10,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rsi
  and    $0xfffffffffffffff0,%rdx
  cmp    %rsi,%rsp
  je     L34
L61:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L61
L34:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  je     L105
  orq    $0x0,-0x8(%rsp,%rdx,1)
L105:
  mov    %rsp,%r9
  mov    %r14d,%ebx
  cmp    $0x1,%r15d
  jle    L2
  cmp    %r15d,%r14d
  jge    L70
  mov    $0x1,%eax
  xor    %r12d,%r12d
  jmp    L88
  nopl   0x0(%rax)
L55:
  add    $0x1,%r12d
  mov    %edx,-0x4(%r9,%rax,4)
  mov    %eax,%r8d
  cmp    %r12d,%r14d
  jle    L26
L91:
  add    $0x1,%rax
  cmp    %r15d,%ebx
  jge    L46
L88:
  movslq %r12d,%rdx
  movslq %ebx,%rsi
  mov    0x0(%r13,%rdx,4),%edx
  mov    0x0(%r13,%rsi,4),%esi
  cmp    %esi,%edx
  jl     L55
  mov    %esi,%edx
  add    $0x1,%ebx
  mov    %eax,%r8d
  mov    %edx,-0x4(%r9,%rax,4)
  cmp    %r12d,%r14d
  jg     L91
L26:
  cmp    %ebx,%r15d
  jle    L60
  lea    -0x1(%r15),%eax
  movslq %r8d,%r8
  mov    %r10,-0x58(%rbp)
  sub    %ebx,%eax
  movslq %ebx,%rbx
  lea    (%r9,%r8,4),%rdi
  mov    %r9,-0x50(%rbp)
  lea    0x4(,%rax,4),%rdx
  lea    0x0(%r13,%rbx,4),%rsi
  call   <memcpy@plt>
  mov    -0x58(%rbp),%r10
  mov    -0x50(%rbp),%r9
L60:
  test   %r15d,%r15d
  jle    L152
  lea    0x0(,%r10,4),%rdx
  mov    %r9,%rsi
  mov    %r13,%rdi
  call   <memcpy@plt>
L152:
  mov    -0x48(%rbp),%rsp
L154:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L86
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L70:
  xor    %r8d,%r8d
  xor    %r12d,%r12d
  nop
L46:
  lea    -0x1(%r14),%r11d
  movslq %r8d,%rax
  mov    %r10,-0x68(%rbp)
  sub    %r12d,%r11d
  lea    (%r9,%rax,4),%rdi
  cmp    %r12d,%r14d
  mov    %r8d,-0x5c(%rbp)
  mov    %r11d,%eax
  mov    %r9,-0x58(%rbp)
  lea    0x4(,%rax,4),%rdx
  mov    $0x4,%eax
  mov    %r11d,-0x50(%rbp)
  cmovle %rax,%rdx
  movslq %r12d,%rax
  lea    0x0(%r13,%rax,4),%rsi
  call   <memcpy@plt>
  mov    -0x50(%rbp),%r11d
  xor    %eax,%eax
  cmp    %r12d,%r14d
  mov    -0x5c(%rbp),%r8d
  mov    -0x58(%rbp),%r9
  cmovle %eax,%r11d
  mov    -0x68(%rbp),%r10
  lea    0x1(%r8,%r11,1),%r8d
  jmp    L26
L2:
  xor    %r8d,%r8d
  jmp    L26
L86:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void ms(int *a, int l) {
  if (l == 1)
    return;
  ms(a, l / 2), ms(&a[l / 2], l - l / 2);
  int i = 0, j = l / 2, k = 0, A[l];
  while ((i < l / 2) && (j < l))
    A[k++] = a[i] < a[j] ? a[i++] : a[j++];
  while (i < l / 2)
    A[k++] = a[i++];
  while (j < l)
    A[k++] = a[j++];
  for (k = 0; k < l; ++k)
    a[k] = A[k];
}
```