#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#define rep(i,s,t) for(register int i=s;i<=t;++i)
using namespace std;
const int N=1e6+11;
char s[N];
int n,u[N],v[N],top,t;
int main(){
	scanf("%s",s+1);
	n=strlen(s+1);
	rep(i,1,n-1)
		if(s[i]!=s[n-i]){
			puts("-1");
			return 0;
		}
	if(s[n]=='1'||s[1]=='0'){
		puts("-1");
		return 0;
	}
	u[1]=n-1,v[1]=n;
	t=n-1;
	top=n-1;
	rep(i,2,n-1)
		if(s[i]=='1')
			u[i]=--t,v[i]=top,top=t;
		else
			u[i]=--t,v[i]=top;
	rep(i,1,n-1)
		printf("%d %d\n",u[i],v[i]);
	return 0;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#define rep(i,s,t) for(register int i=s;i<=t;++i)
using namespace std;
const int N=1e6+11;
char s[N];
int n,u[N],v[N],top,t;
int main(){
	scanf("%s",s+1);
	n=strlen(s+1);
	rep(i,1,n-1)
		if(s[i]!=s[n-i]){
			puts("-1");
			return 0;
		}
	if(s[n]=='1'||s[1]=='0'){
		puts("-1");
		return 0;
	}
	u[1]=n-1,v[1]=n;
	t=n-1;
	top=n-1;
	rep(i,2,n-1)
		if(s[i]=='1')
			u[i]=--t,v[i]=top,top=t;
		else
			u[i]=--t,v[i]=top;
	rep(i,1,n-1)
		printf("%d %d\n",u[i],v[i]);
	return 0;
}
```