#include<bits/stdc++.h>
using namespace std;
const int N=1e5+100;

int n,cnt,u[N],v[N],num;
char s[N];

int main(){
	int i,j,pre,nw;
	scanf("%s",s+1);
	n=strlen(s+1);
	for(i=1;i<=n;++i) s[i]-='0';
	if((!s[1])||(!s[n-1])||(s[n])) {puts("-1");return 0;}
	for(i=2;i<n-i;++i) 
	 if(s[i]!=s[n-i]){puts("-1");return 0;}
	u[++cnt]=1;v[cnt]=2;
	for(pre=1,i=2;i<n;++i){
		for(;i<n && (!s[i]);++i);
		j=i-pre-1;nw=cnt+1;
		for(;j;--j){cnt++;u[cnt]=nw;v[cnt]=cnt+1;}
		u[++cnt]=nw;v[cnt]=cnt+1;
		pre=i;
	}
	for(i=1;i<=cnt;++i) printf("%d %d\n",u[i],v[i]);
}
### Prompt

Please create a solution in Cpp 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 N=1e5+100;

int n,cnt,u[N],v[N],num;
char s[N];

int main(){
	int i,j,pre,nw;
	scanf("%s",s+1);
	n=strlen(s+1);
	for(i=1;i<=n;++i) s[i]-='0';
	if((!s[1])||(!s[n-1])||(s[n])) {puts("-1");return 0;}
	for(i=2;i<n-i;++i) 
	 if(s[i]!=s[n-i]){puts("-1");return 0;}
	u[++cnt]=1;v[cnt]=2;
	for(pre=1,i=2;i<n;++i){
		for(;i<n && (!s[i]);++i);
		j=i-pre-1;nw=cnt+1;
		for(;j;--j){cnt++;u[cnt]=nw;v[cnt]=cnt+1;}
		u[++cnt]=nw;v[cnt]=cnt+1;
		pre=i;
	}
	for(i=1;i<=cnt;++i) printf("%d %d\n",u[i],v[i]);
}
```