#include <bits/stdc++.h>
using namespace std;
char s[111111];
int x[111111],cnt;
int main(){
	scanf("%s",s+1); int n=strlen(s+1);
//	puts("WTF");
	if (s[1]=='0') {return puts("-1"),0;}
	if (s[n]=='1') {return puts("-1"),0;}
	for (int i=1,j=n-1;i<=j;i++,j--){
		if (s[i]!=s[j]) return puts("-1"),0;
		if (s[i]=='1') x[++cnt]=i;
	}
	x[++cnt]=n;
	for (int i=1;i<cnt;i++) {
		printf("%d %d\n",i,i+1);
	} 
	int id=cnt;
	for (int i=2;i<=cnt;i++){
		int sz=x[i]-x[i-1]-1;
		for (int j=0;j<sz;j++) printf("%d %d\n",i,++id);
	}
	return 0;
}
	
	
	
### Prompt

Construct a CPP code solution to the problem outlined:

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;
char s[111111];
int x[111111],cnt;
int main(){
	scanf("%s",s+1); int n=strlen(s+1);
//	puts("WTF");
	if (s[1]=='0') {return puts("-1"),0;}
	if (s[n]=='1') {return puts("-1"),0;}
	for (int i=1,j=n-1;i<=j;i++,j--){
		if (s[i]!=s[j]) return puts("-1"),0;
		if (s[i]=='1') x[++cnt]=i;
	}
	x[++cnt]=n;
	for (int i=1;i<cnt;i++) {
		printf("%d %d\n",i,i+1);
	} 
	int id=cnt;
	for (int i=2;i<=cnt;i++){
		int sz=x[i]-x[i-1]-1;
		for (int j=0;j<sz;j++) printf("%d %d\n",i,++id);
	}
	return 0;
}
	
	
	
```