#include<bits/stdc++.h>
using namespace std ;
const int maxn = 1e5 + 10 ;
string s ;
int main()
{
	cin >> s ;
	bool b  = 1 ;
	if ( s[0] != '1' || s[s.size()-1] != '0' ) 
		b = false ;
	for ( int i = 0; i< s.size()-1 ; i++)
		if ( s[i] != s[s.size()-2-i] ) 
			b = false ;
	if ( !b ) 
		cout << -1<< endl;
	else
	{
		int x = 1;
		for ( int i = 0 ; i<s.size()-1 ; i++)
		{
			cout << x << " " << i+2 << "\n" ;
		if ( s[i] == '1' ) 
			x = i+ 2 ;
		}
	}

}
### Prompt

Please create a solution in Cpp to 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 ;
const int maxn = 1e5 + 10 ;
string s ;
int main()
{
	cin >> s ;
	bool b  = 1 ;
	if ( s[0] != '1' || s[s.size()-1] != '0' ) 
		b = false ;
	for ( int i = 0; i< s.size()-1 ; i++)
		if ( s[i] != s[s.size()-2-i] ) 
			b = false ;
	if ( !b ) 
		cout << -1<< endl;
	else
	{
		int x = 1;
		for ( int i = 0 ; i<s.size()-1 ; i++)
		{
			cout << x << " " << i+2 << "\n" ;
		if ( s[i] == '1' ) 
			x = i+ 2 ;
		}
	}

}
```