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

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  movq   $0x0,0x20(%rsp)
  movq   $0x0,0x28(%rsp)
  movq   $0x0,0x30(%rsp)
  lea    0x10(%rsp),%rbx
  mov    $0x0,%r13d
  lea    D42(%rip),%r12
L39:
  mov    %rbx,%rbp
  mov    %rbx,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  test   %rax,%rax
  jle    L197
  mov    (%rbx),%eax
  cmp    $0xa,%eax
  je     L197
  sub    $0x30,%eax
  mov    %eax,0x0(%rbp)
  add    $0x1,%r13d
  add    $0x4,%rbx
  cmp    $0xa,%r13d
  jne    L39
L197:
  movslq %r13d,%rax
  movl   $0x0,0x10(%rsp,%rax,4)
  movl   $0x0,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  lea    D6(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    $0x0,%r14d
  mov    $0x0,%r12d
  lea    -0x1(%r13),%ebp
  lea    D6(%rip),%r15
  jmp    L95
L85:
  movslq (%rdx),%rsi
  sub    %rsi,%rcx
L159:
  add    $0x1,%eax
  add    $0x4,%rdx
  cmp    %ebp,%eax
  je     L164
L98:
  bt     %eax,%ebx
  jae    L85
  movslq (%rdx),%rsi
  add    %rsi,%rcx
  jmp    L159
L25:
  cmp    0xc(%rsp),%eax
  jne    L67
  mov    $0x0,%r12d
  cmp    $0x1,%r13d
  je     L193
L183:
  mov    0xc(%rsp),%esi
  lea    D19(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%r12d
  jmp    L67
L89:
  mov    $0x0,%r12d
  jmp    L193
L59:
  mov    $0x2d,%edi
  call   <putchar@plt>
L106:
  add    $0x1,%r12
  cmp    %r12d,%r13d
  jle    L183
L193:
  mov    0x10(%rsp,%r12,4),%esi
  mov    %r15,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  bt     %r12d,%ebx
  jae    L59
  cmp    %r12d,%ebp
  jle    L59
  mov    $0x2b,%edi
  call   <putchar@plt>
  jmp    L106
L154:
  test   %r12b,%r12b
  je     L107
L143:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L144
  mov    $0x0,%eax
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L107:
  lea    D26(%rip),%rdi
  call   <puts@plt>
  jmp    L143
L164:
  movslq 0xc(%rsp),%rax
  cmp    %rax,%rcx
  je     L89
L67:
  add    $0x1,%r14
L95:
  mov    %r14d,%ebx
  mov    %ebp,%esi
  mov    $0x2,%edi
  call   <powl@plt>
  cmp    %r14,%rax
  jl     L154
  mov    0x10(%rsp),%eax
  movslq %eax,%rcx
  cmp    $0x1,%r13d
  jle    L25
  lea    0x14(%rsp),%rdx
  mov    $0x0,%eax
  jmp    L98
L144:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D42
data_type: string
name: disassemble_data
arguments: tag: D6
data_type: string
name: disassemble_data
arguments: tag: D19
data_type: string
name: disassemble_data
arguments: tag: D26
data_type: string

tool: D42: %c

tool: D6: %d

tool: D19: =%d


tool: D26: -1

assistant: ```c
int main() {
  int a[10] = {0};
  int i;
  for (i = 0; i < 10 && scanf("%c", (char *)&a[i]) > 0; ++i) {
    if (a[i] == 10)
      break;
    else {
      a[i] = a[i] - '0';
    }
  }
  a[i] = 0;
  char ima = 0;
  int rez = 0;
  scanf("%d", &rez);
  long sum = 0;
  for (int j = 0; j <= (long int)powl(2, i - 1); ++j) {
    sum = 0;
    sum = a[0];
    for (int k = 1; k <= i - 1; ++k) {
      if (((j >> (k - 1)) & 1)) {
        sum = sum + a[k];
      } else {
        sum = sum - a[k];
      }
    }
    if ((long int)rez == sum) {
      ima = 1;
      for (int k = 0; k < i; ++k) {
        printf("%d", a[k]);
        if (((j >> k) & 1) && k < i - 1) {
          printf("+");
        } else {
          printf("-");
        }
      }
      printf("\b=%d\n", rez);
    }
  }
  if (ima == 0)
    printf("-1\n");
  return 0;
}
```