#include<bits/stdc++.h>
using namespace std;
string s;

int main()
{
	cin>>s;
	int n=s.size();
	if(s[n-1]=='1'||s[0]=='0')
	{
		puts("-1");
		return 0;
	}
	for(int i=0;i<n-1;i++)
	{
		if(s[i]!=s[n-i-2])
		{
			puts("-1");
			return 0; 
		}
	}
	int pre=1;
	for(int i=1;i<=n-1;i++)
	{
		if(s[i-1]=='1')
		{
			cout<<i+1<<' '<<pre<<endl;
			pre=i+1;
		}
		else
		{
//			cout<<pre<<endl;
			cout<<i+1<<' '<<pre<<endl; 
		}
	}
	//cout<<n-1<<ends<<n<<endl;
	return 0;
} 
### Prompt

In cpp, your task is to solve 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<bits/stdc++.h>
using namespace std;
string s;

int main()
{
	cin>>s;
	int n=s.size();
	if(s[n-1]=='1'||s[0]=='0')
	{
		puts("-1");
		return 0;
	}
	for(int i=0;i<n-1;i++)
	{
		if(s[i]!=s[n-i-2])
		{
			puts("-1");
			return 0; 
		}
	}
	int pre=1;
	for(int i=1;i<=n-1;i++)
	{
		if(s[i-1]=='1')
		{
			cout<<i+1<<' '<<pre<<endl;
			pre=i+1;
		}
		else
		{
//			cout<<pre<<endl;
			cout<<i+1<<' '<<pre<<endl; 
		}
	}
	//cout<<n-1<<ends<<n<<endl;
	return 0;
} 
```