#include<cstdio>
#define MAXN 100006
#include<cstring>
#include<algorithm>
using namespace std;
int n,a[MAXN],fa[MAXN],root;
char s[MAXN];
int main()
{
	scanf("%s",s+1);
	n=strlen(s+1);
	for(int i=1;i<=n;i++)
		a[i]=s[i]-'0';
	if(a[1]==0)
	{
		printf("%d",-1);
		return 0;
	}
	for(int i=1;i<=n;i++)
		if(a[i]!=a[n-i])
		{
			printf("-1");
			return 0;
		}
	int m=n/2;
	for(int i=m+1;i<n;i++)
		fa[i]=n;
	root=n;
	for(int i=m;i>=1;i--)
	{
		fa[i]=root;
		if(a[i]==1) root=i;
	}
	for(int i=1;i<n;i++)
		printf("%d %d\n",i,fa[i]);
}
### Prompt

Develop a solution in Cpp 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>
#define MAXN 100006
#include<cstring>
#include<algorithm>
using namespace std;
int n,a[MAXN],fa[MAXN],root;
char s[MAXN];
int main()
{
	scanf("%s",s+1);
	n=strlen(s+1);
	for(int i=1;i<=n;i++)
		a[i]=s[i]-'0';
	if(a[1]==0)
	{
		printf("%d",-1);
		return 0;
	}
	for(int i=1;i<=n;i++)
		if(a[i]!=a[n-i])
		{
			printf("-1");
			return 0;
		}
	int m=n/2;
	for(int i=m+1;i<n;i++)
		fa[i]=n;
	root=n;
	for(int i=m;i>=1;i--)
	{
		fa[i]=root;
		if(a[i]==1) root=i;
	}
	for(int i=1;i<n;i++)
		printf("%d %d\n",i,fa[i]);
}
```