#include <cstdio>
#include <cstring>

char s[100010];
int len;

void solve()
{
	int i;
	int cur=1;
	len = strlen(s);
	if(s[len-1] == '1' || s[0] == '0'){
		printf("-1\n");
		return ;
	}
	for(i=0; i<(len-1)/2; i++){
		if(s[i] != s[len-2-i]){
			printf("-1\n");
			return ;
		}
	}
	int p=1;
	for(i=0; i<len-1; i++){
		if(s[i] == '0'){
			cur ++;
			printf("%d %d\n", p, cur);
		}
		else{
			cur ++;
			printf("%d %d\n", p, cur);
			p = cur;
		}
	}
}

int main()
{
	scanf("%s", s);
	solve();
	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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 <cstring>

char s[100010];
int len;

void solve()
{
	int i;
	int cur=1;
	len = strlen(s);
	if(s[len-1] == '1' || s[0] == '0'){
		printf("-1\n");
		return ;
	}
	for(i=0; i<(len-1)/2; i++){
		if(s[i] != s[len-2-i]){
			printf("-1\n");
			return ;
		}
	}
	int p=1;
	for(i=0; i<len-1; i++){
		if(s[i] == '0'){
			cur ++;
			printf("%d %d\n", p, cur);
		}
		else{
			cur ++;
			printf("%d %d\n", p, cur);
			p = cur;
		}
	}
}

int main()
{
	scanf("%s", s);
	solve();
	return 0;
}
```