#include <cstdio>
#include <cstring>

const int maxn=100000;

char s[maxn+10];
int len;

int main()
{
  scanf("%s",s+1);
  len=strlen(s+1);
  if((s[len]!='0')||(s[1]!='1'))
    {
      puts("-1");
      return 0;
    }
  for(int i=1; i<=len-i; ++i)
    {
      if(s[i]!=s[len-i])
        {
          puts("-1");
          return 0;
        }
    }
  int past=1;
  for(int i=1; i<len; ++i)
    {
      printf("%d %d\n",past,i+1);
      if(s[i]=='1')
        {
          past=i+1;
        }
    }
  return 0;
}

### Prompt

Create a solution in CPP for the following problem:

You are given a string s of length n. Does a tree with n vertices that satisfies the following conditions exist?

* The vertices are numbered 1,2,..., n.
* The edges are numbered 1,2,..., n-1, and Edge i connects Vertex u_i and v_i.
* If the i-th character in s is `1`, we can have a connected component of size i by removing one edge from the tree.
* If the i-th character in s is `0`, we cannot have a connected component of size i by removing any one edge from the tree.



If such a tree exists, construct one such tree.

Constraints

* 2 \leq n \leq 10^5
* s is a string of length n consisting of `0` and `1`.

Input

Input is given from Standard Input in the following format:


s


Output

If a tree with n vertices that satisfies the conditions does not exist, print `-1`.

If a tree with n vertices that satisfies the conditions exist, print n-1 lines. The i-th line should contain u_i and v_i with a space in between. If there are multiple trees that satisfy the conditions, any such tree will be accepted.

Examples

Input

1111


Output

-1


Input

1110


Output

1 2
2 3
3 4


Input

1010


Output

1 2
1 3
1 4

### Response

```cpp
#include <cstdio>
#include <cstring>

const int maxn=100000;

char s[maxn+10];
int len;

int main()
{
  scanf("%s",s+1);
  len=strlen(s+1);
  if((s[len]!='0')||(s[1]!='1'))
    {
      puts("-1");
      return 0;
    }
  for(int i=1; i<=len-i; ++i)
    {
      if(s[i]!=s[len-i])
        {
          puts("-1");
          return 0;
        }
    }
  int past=1;
  for(int i=1; i<len; ++i)
    {
      printf("%d %d\n",past,i+1);
      if(s[i]=='1')
        {
          past=i+1;
        }
    }
  return 0;
}

```