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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xa20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xa20(%rbp)
  jmp    L55
L68:
  movl   $0x0,-0xa1c(%rbp)
  jmp    L64
L49:
  mov    -0xa1c(%rbp),%eax
  movslq %eax,%rcx
  mov    -0xa20(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  add    %rdx,%rax
  add    %rax,%rax
  add    %rbp,%rax
  add    %rcx,%rax
  sub    $0xa10,%rax
  movb   $0x20,(%rax)
  addl   $0x1,-0xa1c(%rbp)
L64:
  cmpl   $0x31,-0xa1c(%rbp)
  jle    L49
  pxor   %xmm0,%xmm0
  cvtsi2ssl -0xa20(%rbp),%xmm0
  movss  %xmm0,-0xa18(%rbp)
  movss  -0xa18(%rbp),%xmm0
  movss  D21(%rip),%xmm1
  divss  %xmm1,%xmm0
  pxor   %xmm2,%xmm2
  cvtss2sd %xmm0,%xmm2
  movq   %xmm2,%rax
  movq   %rax,%xmm0
  mov    $0x1,%eax
  call   <f@plt>
  imul   $0x32,%eax,%eax
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  movss  %xmm0,-0xa18(%rbp)
  movss  -0xa18(%rbp),%xmm0
  movss  D20(%rip),%xmm1
  comiss %xmm1,%xmm0
  jbe    L10
  movss  D21(%rip),%xmm0
  comiss -0xa18(%rbp),%xmm0
  jbe    L10
  pxor   %xmm3,%xmm3
  cvtss2sd -0xa18(%rbp),%xmm3
  movq   %xmm3,%rax
  movq   %rax,%xmm0
  mov    $0x1,%eax
  call   <C@plt>
  mov    %eax,%edx
  mov    $0x31,%eax
  sub    %edx,%eax
  mov    %eax,-0xa14(%rbp)
  mov    -0xa14(%rbp),%eax
  movslq %eax,%rcx
  mov    -0xa20(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  add    %rdx,%rax
  add    %rax,%rax
  add    %rbp,%rax
  add    %rcx,%rax
  sub    $0xa10,%rax
  movb   $0x91,(%rax)
L10:
  addl   $0x1,-0xa20(%rbp)
L55:
  cmpl   $0x31,-0xa20(%rbp)
  jle    L68
  movl   $0x0,-0xa20(%rbp)
  jmp    L86
L51:
  movl   $0x0,-0xa1c(%rbp)
  jmp    L90
L66:
  mov    -0xa20(%rbp),%eax
  movslq %eax,%rcx
  mov    -0xa1c(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  add    %rdx,%rax
  add    %rax,%rax
  add    %rbp,%rax
  add    %rcx,%rax
  sub    $0xa10,%rax
  movzbl (%rax),%edx
  mov    -0xa1c(%rbp),%eax
  cltq
  mov    %dl,-0x40(%rbp,%rax,1)
  addl   $0x1,-0xa1c(%rbp)
L90:
  cmpl   $0x31,-0xa1c(%rbp)
  jle    L66
  lea    -0x40(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <print@plt>
  mov    $0xa,%edi
  mov    $0x0,%eax
  call   <print@plt>
  addl   $0x1,-0xa20(%rbp)
L86:
  cmpl   $0x31,-0xa20(%rbp)
  jle    L51
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L41
  call   <__stack_chk_fail@plt>
L41:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D21
data_type: byte[4]
name: retrieve_data
arguments: data_tag: D20
data_type: byte[4]

tool: {"D21": [0x00,0x00,0x48,0x42]}

tool: {"D20": [0x00,0x00,0x80,0xbf]}

assistant: ```c
void main() {
  char ekr[50][50], t[50];
  int i, j, y;
  float x;
  for (i = 0; i < 50; i++) {
    for (j = 0; j < 50; j++)
      ekr[i][j] = ' ';
    x = i;
    x = f(x / 50) * 50;
    if ((x > -1) && (x < 50)) {
      y = 50 - 1 - C(x);
      ekr[i][y] = 'ё';
    }
  }
  for (i = 0; i < 50; i++) {
    for (j = 0; j < 50; j++)
      t[j] = ekr[j][i];
    print(t);
    print('\n');
  }
}
```