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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  mov    $0x0,%r14d
  lea    -0x3c(%rbp),%rax
  mov    %rax,-0x48(%rbp)
  jmp    L140
L36:
  mov    %rsi,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L180
  orq    $0x0,-0x8(%rsp,%rax,1)
L180:
  mov    %rsp,%r13
  test   %edi,%edi
  jle    L62
  mov    %r13,%rax
  lea    -0x2(%rdx),%edx
  lea    0x4(%r13,%rdx,4),%rdx
L234:
  movl   $0x0,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L234
  test   %r8d,%r8d
  jle    L10
L152:
  lea    -0x1(%r9),%esi
  lea    0x8(%r13,%rsi,4),%rdx
  mov    $0x1,%ecx
  not    %rsi
  shl    $0x2,%rsi
  jmp    L6
L62:
  test   %r8d,%r8d
  jg     L152
L201:
  test   %r14d,%r14d
  jne    L101
L40:
  add    $0x1,%r14d
  mov    %r15,%rsp
L140:
  lea    -0x40(%rbp),%rsi
  mov    -0x48(%rbp),%rdx
  lea    D3(%rip),%rdi
  call   <scanf@plt>
  cmp    $0x2,%eax
  jne    L3
  mov    %rsp,%r15
  mov    -0x40(%rbp),%r8d
  mov    -0x3c(%rbp),%r9d
  lea    (%r8,%r9,1),%edx
  lea    -0x1(%rdx),%edi
  movslq %edi,%rax
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rsi
  and    $0xfffffffffffffff0,%rsi
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rcx
  sub    %rax,%rcx
L23:
  cmp    %rcx,%rsp
  je     L36
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L23
L124:
  addl   $0x1,-0x4(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L124
L198:
  add    $0x1,%ecx
  add    $0x4,%rdx
  cmp    %ecx,%r8d
  jl     L86
L6:
  lea    (%rsi,%rdx,1),%rax
  test   %r9d,%r9d
  jg     L124
  jmp    L198
L86:
  test   %edi,%edi
  jle    L201
L10:
  mov    $0x0,%eax
  mov    $0x0,%r12d
L55:
  mov    0x0(%r13,%rax,4),%edx
  cmp    %edx,%r12d
  cmovl  %edx,%r12d
  add    $0x1,%rax
  cmp    %eax,%edi
  jg     L55
  test   %r14d,%r14d
  jne    L226
L139:
  mov    $0x0,%ebx
  jmp    L231
L101:
  mov    $0x0,%r12d
L226:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0x3c(%rbp),%eax
  add    -0x40(%rbp),%eax
  cmp    $0x1,%eax
  jg     L139
  jmp    L40
L142:
  add    $0x1,%rbx
  mov    -0x3c(%rbp),%eax
  add    -0x40(%rbp),%eax
  sub    $0x1,%eax
  cmp    %ebx,%eax
  jle    L40
L231:
  cmp    %r12d,0x0(%r13,%rbx,4)
  jne    L142
  lea    0x2(%rbx),%esi
  lea    D6(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L142
L3:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L161
  mov    $0x0,%eax
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L161:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D3
type_infered: string
name: extract_symbolic_data
arguments: tag: D6
type_infered: string

tool: D3: %d %d

tool: D6: %d


assistant: ```c
int main() {
  int d1, d2, mx, i, j, t = 0;
  while (scanf("%d %d", &d1, &d2) == 2) {
    mx = 0;
    int temp[d1 + d2 - 1];
    for (i = 0; i < d1 + d2 - 1; i++)
      temp[i] = 0;
    for (j = 1; j <= d1; j++)
      for (i = 1; i <= d2; i++) {
        temp[j + i - 2] += 1;
      }
    for (i = 0; i < d1 + d2 - 1; i++)
      if (mx < temp[i])
        mx = temp[i];
    if (t != 0)
      printf("\n");
    for (i = 0; i < d1 + d2 - 1; i++)
      if (mx == temp[i])
        printf("%d\n", i + 2);
    t++;
  }
  return 0;
}
```