#include <bits/stdc++.h>

using namespace std;

int n;
char s[500010];
int main()
{
	scanf("%s",s+1);
	n=strlen(s+1);
	if(s[1]=='0'||s[n]=='1')
	{
		puts("-1");
		return 0;
	}
	for(int i=2;i<=n-1;i++)
		if(s[i]!=s[n-i])
		{
			puts("-1");
			return 0;
		}
	int tmp=2,pr=1;
	while(tmp<=n)
	{
		int st=tmp; 
		printf("%d %d\n",pr,st);
		while(s[tmp]=='0') tmp++;
		for(int j=st+1;j<=min(n,tmp);j++)
			printf("%d %d\n",st,j);
		tmp++; pr=st;
	}
	return 0;	
}
### Prompt

Your task is to create a Cpp solution 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;

int n;
char s[500010];
int main()
{
	scanf("%s",s+1);
	n=strlen(s+1);
	if(s[1]=='0'||s[n]=='1')
	{
		puts("-1");
		return 0;
	}
	for(int i=2;i<=n-1;i++)
		if(s[i]!=s[n-i])
		{
			puts("-1");
			return 0;
		}
	int tmp=2,pr=1;
	while(tmp<=n)
	{
		int st=tmp; 
		printf("%d %d\n",pr,st);
		while(s[tmp]=='0') tmp++;
		for(int j=st+1;j<=min(n,tmp);j++)
			printf("%d %d\n",st,j);
		tmp++; pr=st;
	}
	return 0;	
}
```