#include<bits/stdc++.h>
using namespace std;
string s;
int main(){
	cin>>s;
	int n=s.length();
	bool c=true;
	if (s[n-1]=='1')c=false;
	if (s[0]=='0')c=false;
	for (int i=1;i<=n/2;i++)if (s[i-1]!=s[n-i-1])c=false;
	if (!c){cout<<"-1"<<endl;}
	else{
		int no=1;
		for (int i=1;i<=n/2;i++){
			if (s[i-1]=='0'){
				cout<<no<<' '<<i+1<<endl;
			}else{
				cout<<no<<' '<<i+1<<endl;no=i+1;
			}
		}
		for (int i=n/2+2;i<=n;i++)cout<<no<<' '<<i<<endl;
	}
	return 0;
}
### Prompt

Develop a solution in cpp to the problem described below:

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.length();
	bool c=true;
	if (s[n-1]=='1')c=false;
	if (s[0]=='0')c=false;
	for (int i=1;i<=n/2;i++)if (s[i-1]!=s[n-i-1])c=false;
	if (!c){cout<<"-1"<<endl;}
	else{
		int no=1;
		for (int i=1;i<=n/2;i++){
			if (s[i-1]=='0'){
				cout<<no<<' '<<i+1<<endl;
			}else{
				cout<<no<<' '<<i+1<<endl;no=i+1;
			}
		}
		for (int i=n/2+2;i<=n;i++)cout<<no<<' '<<i<<endl;
	}
	return 0;
}
```