#include <iostream>
using namespace std;
int main(int argc, char** argv) {
	string a;
	cin >> a;
	int n=a.size();
	a='0'+a;
	for(int i=0;i<=n;i++)
	{
		if(a[i]^a[n-i])
		{
			puts("-1");
			return 0;
		}
	}
	if(a[1]=='0')
	{
		puts("-1");
		return 0;
	}
	int cnt=1,rt=1;
	for(int i=1;i<=n/2;i++)
	{
		if(a[i]=='0') cout << rt << " " << cnt+1 << "\n";
		else
		{
			cout << rt << ' ' << cnt+1 << "\n";
			rt=cnt+1;
		}
		++cnt;
	}
	for(int i=n/2+1;i<n;i++)
	{ 
		++cnt;
		cout << rt << " " << cnt << "\n";
	}
	return 0;
}
### Prompt

Please provide a cpp coded solution 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 <iostream>
using namespace std;
int main(int argc, char** argv) {
	string a;
	cin >> a;
	int n=a.size();
	a='0'+a;
	for(int i=0;i<=n;i++)
	{
		if(a[i]^a[n-i])
		{
			puts("-1");
			return 0;
		}
	}
	if(a[1]=='0')
	{
		puts("-1");
		return 0;
	}
	int cnt=1,rt=1;
	for(int i=1;i<=n/2;i++)
	{
		if(a[i]=='0') cout << rt << " " << cnt+1 << "\n";
		else
		{
			cout << rt << ' ' << cnt+1 << "\n";
			rt=cnt+1;
		}
		++cnt;
	}
	for(int i=n/2+1;i<n;i++)
	{ 
		++cnt;
		cout << rt << " " << cnt << "\n";
	}
	return 0;
}
```