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

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  lea    D15(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x340,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x338(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rbx
  nopw   0x0(%rax,%rax,1)
L8:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  mov    %eax,%r14d
  and    $0x1,%r14d
  je     L8
  test   %eax,%eax
  jle    L102
  lea    0x10(%rsp),%rbp
  mov    %r12,%rdi
  lea    0x14(%rsp),%r13
  mov    %rbp,%rsi
  mov    %r13,%rbx
  call   <scanf@plt>
L53:
  mov    0xc(%rsp),%esi
  cmp    %r14d,%esi
  jle    L110
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    (%rbx),%edx
  mov    %rbp,%rax
  mov    $0x1,%ecx
  jmp    L54
  nopl   0x0(%rax)
L143:
  add    $0x4,%rax
  cmp    %rbx,%rax
  je     L29
L54:
  cmp    (%rax),%edx
  jne    L143
  cmp    $0x2,%ecx
  je     L102
  add    $0x4,%rax
  mov    $0x2,%ecx
  cmp    %rbx,%rax
  jne    L54
L29:
  add    $0x1,%r14d
  add    $0x4,%rbx
  jmp    L53
L210:
  mov    %r8d,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  nopl   0x0(%rax)
L102:
  mov    0x338(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L121
  add    $0x340,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L110:
  test   %esi,%esi
  jle    L102
  lea    -0x1(%rsi),%eax
  mov    %esi,%r10d
  mov    %rbp,%rdi
  mov    %rax,%r11
  lea    0x0(%r13,%rax,4),%rbx
  mov    %esi,%eax
  and    $0xfffffffc,%r10d
  shr    $0x2,%eax
  sub    $0x1,%eax
  shl    $0x4,%rax
  lea    0x20(%rsp,%rax,1),%rcx
  nopl   0x0(%rax,%rax,1)
L32:
  mov    (%rdi),%r8d
  cmp    $0x2,%r11d
  jbe    L207
  movd   %r8d,%xmm3
  mov    %rbp,%rax
  pxor   %xmm0,%xmm0
  pshufd $0x0,%xmm3,%xmm2
  xchg   %ax,%ax
L9:
  movdqa %xmm2,%xmm1
  add    $0x10,%rax
  pcmpeqd -0x10(%rax),%xmm1
  psubd  %xmm1,%xmm0
  cmp    %rax,%rcx
  jne    L9
  movdqa %xmm0,%xmm1
  mov    %r10d,%edx
  psrldq $0x8,%xmm1
  paddd  %xmm1,%xmm0
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%eax
  cmp    %esi,%r10d
  je     L61
L119:
  movslq %edx,%r9
  cmp    0x10(%rsp,%r9,4),%r8d
  jne    L194
  add    $0x1,%eax
L194:
  lea    0x1(%rdx),%r9d
  cmp    %r9d,%esi
  jle    L61
  movslq %r9d,%r9
  cmp    0x10(%rsp,%r9,4),%r8d
  jne    L200
  add    $0x1,%eax
L200:
  add    $0x2,%edx
  cmp    %edx,%esi
  jle    L61
  movslq %edx,%rdx
  cmp    0x10(%rsp,%rdx,4),%r8d
  jne    L61
  add    $0x1,%eax
  nopl   0x0(%rax,%rax,1)
L61:
  cmp    $0x1,%eax
  je     L210
  add    $0x4,%rdi
  cmp    %rbx,%rdi
  jne    L32
  jmp    L102
L207:
  xor    %edx,%edx
  xor    %eax,%eax
  jmp    L119
L121:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D15
data_type: string
name: read_data
arguments: tag: D2
data_type: string

tool: D15: %d

tool: D2: %d 


assistant: ```c
int main() {
  int number, counter, check, counterdeer, deer[200];
  do {
    scanf("%d", &number);
  } while (number % 2 == 0);
  for (counter = 0; counter < number; counter++) {
    scanf("%d", &deer[counter]);
    for (counterdeer = 0, check = 1; counterdeer < counter; counterdeer++) {
      if (deer[counter] == deer[counterdeer]) {
        check++;
      }
      if (check > 2) {
        return 0;
      }
    }
  }
  for (counter = 0; counter < number; counter++) {
    check = 0;
    for (counterdeer = 0; counterdeer < number; counterdeer++) {
      if (deer[counter] == deer[counterdeer]) {
        check++;
      }
    }
    if (check == 1) {
      printf("%d \n", deer[counter]);
      break;
    }
  }
}
```