#include<stdio.h>
#include<string.h>
#define fo(i,a,b) for(int i=a;i<=b;i++)
char s[110000];
int n,x[110000],y[110000],now,vt,et;
int main(){
	scanf("%s",s+1);
	n=strlen(s+1);
	if (s[1]!='1'||s[n]!='0'){
		printf("-1\n");
		return 0;
	}
	now=1;vt=1;
	fo(i,1,n-1){
		if (s[i]!=s[n-i]){
			printf("-1\n");
			return 0;			
		}
		++vt;
		x[++et]=now;
		y[et]=vt;
		if (s[i]=='1') now=vt;
	}
	fo(i,1,et) printf("%d %d\n",x[i],y[i]);
	return 0;
} 
### Prompt

In cpp, your task is to solve 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<stdio.h>
#include<string.h>
#define fo(i,a,b) for(int i=a;i<=b;i++)
char s[110000];
int n,x[110000],y[110000],now,vt,et;
int main(){
	scanf("%s",s+1);
	n=strlen(s+1);
	if (s[1]!='1'||s[n]!='0'){
		printf("-1\n");
		return 0;
	}
	now=1;vt=1;
	fo(i,1,n-1){
		if (s[i]!=s[n-i]){
			printf("-1\n");
			return 0;			
		}
		++vt;
		x[++et]=now;
		y[et]=vt;
		if (s[i]=='1') now=vt;
	}
	fo(i,1,et) printf("%d %d\n",x[i],y[i]);
	return 0;
} 
```