#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>

using namespace std;

const int M = 1000001;
int n,m,k,a[M],d[M];
char c[M];
int main()
{
	scanf("\n%s",c+1);
	n=strlen(c+1);
	for(int i=1;i<=n;i++) a[i]=c[i]-'0';
	if(!a[1] || a[n]) 
	{
		printf("-1");
		return 0;
	}
	int m=n/2;
	for(int i=1;i<=m;i++) if(a[i]!=a[n-i]) 
	{
		printf("-1");
		return 0;
	}
	k=2;
	printf("1 2\n");
	for(int i=2;i<=m;i++)
	{
		printf("%d %d\n",i+1,k);
		if(a[i]) k=i+1;
	}
	for(int i=m+2;i<=n;i++) printf("%d %d\n",i,k);
}
### Prompt

Please formulate a Cpp solution 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<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>

using namespace std;

const int M = 1000001;
int n,m,k,a[M],d[M];
char c[M];
int main()
{
	scanf("\n%s",c+1);
	n=strlen(c+1);
	for(int i=1;i<=n;i++) a[i]=c[i]-'0';
	if(!a[1] || a[n]) 
	{
		printf("-1");
		return 0;
	}
	int m=n/2;
	for(int i=1;i<=m;i++) if(a[i]!=a[n-i]) 
	{
		printf("-1");
		return 0;
	}
	k=2;
	printf("1 2\n");
	for(int i=2;i<=m;i++)
	{
		printf("%d %d\n",i+1,k);
		if(a[i]) k=i+1;
	}
	for(int i=m+2;i<=n;i++) printf("%d %d\n",i,k);
}
```