#include<bits/stdc++.h>
using namespace std;
const int N=200005;
int m=0,l;
char s[200005];
int main(){
    scanf("%s",s);
    l=strlen(s) ;
    if(s[0]=='0'){
    	puts("-1");
		return 0;
	}
    if(s[l-1]=='1'){
    	puts("-1");
		return 0;
	}
    for(int i=0;i<l-1;i++)
        if(s[i]!=s[l-i-2]){
    		puts("-1");
			return 0;
		}
    for(int i=0;i<l-1;i++){
        printf("%d %d\n",m+1,i+2);
        if(s[i]=='1')m=i+1;
    }
    return 0 ;
}
### Prompt

Create a solution in CPP for 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 N=200005;
int m=0,l;
char s[200005];
int main(){
    scanf("%s",s);
    l=strlen(s) ;
    if(s[0]=='0'){
    	puts("-1");
		return 0;
	}
    if(s[l-1]=='1'){
    	puts("-1");
		return 0;
	}
    for(int i=0;i<l-1;i++)
        if(s[i]!=s[l-i-2]){
    		puts("-1");
			return 0;
		}
    for(int i=0;i<l-1;i++){
        printf("%d %d\n",m+1,i+2);
        if(s[i]=='1')m=i+1;
    }
    return 0 ;
}
```