#include<bits/stdc++.h>

using namespace std;

const int maxn=1e5+10;

int n;
int ne[maxn],f[maxn];
char ch[maxn];

int main()
{
	scanf("%s",ch+1);
	n=strlen(ch+1);
	
	if(ch[1]=='0'||ch[n]=='1'){
		printf("-1");
		return 0;
	}
	
	int la=0;
	for(int i=1;2*i<=n;i++){
		if(ch[i]=='1')ne[la]=i,la=i;
		if(ch[i]!=ch[n-i]){
			printf("-1");
			return 0;
		}
	}
	
	for(int i=ne[0];ne[i];i=ne[i]){
		f[i]=ne[i];
		for(int j=i+1;j<ne[i];j++)f[j]=ne[i];
	}
	
	for(int i=la;i<n;i++)f[i]=n;
	
	for(int i=1;i<n;i++)
		printf("%d %d\n",i,f[i]);
	
	return 0;
}
### Prompt

Generate 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;

const int maxn=1e5+10;

int n;
int ne[maxn],f[maxn];
char ch[maxn];

int main()
{
	scanf("%s",ch+1);
	n=strlen(ch+1);
	
	if(ch[1]=='0'||ch[n]=='1'){
		printf("-1");
		return 0;
	}
	
	int la=0;
	for(int i=1;2*i<=n;i++){
		if(ch[i]=='1')ne[la]=i,la=i;
		if(ch[i]!=ch[n-i]){
			printf("-1");
			return 0;
		}
	}
	
	for(int i=ne[0];ne[i];i=ne[i]){
		f[i]=ne[i];
		for(int j=i+1;j<ne[i];j++)f[j]=ne[i];
	}
	
	for(int i=la;i<n;i++)f[i]=n;
	
	for(int i=1;i<n;i++)
		printf("%d %d\n",i,f[i]);
	
	return 0;
}
```