#include <stdio.h>
#include <vector>
using namespace std;

int N; char S[100100];

int main()
{
	scanf ("%s",S+1);
	while (S[N+1]) N++;

	if (S[N] == '1' || S[1] == '0'){
		puts("-1");
		return 0;
	}

	for (int i=1;i<N;i++) if (S[i] != S[N-i]){
		puts("-1");
		return 0;
	}

	vector<int> last;
	last.push_back(1);
	for (int i=2;i<=N;i++){
		if (S[i] == '1'){
			for (auto &x : last) printf ("%d %d\n",x,i);
			last.clear();
		}
		last.push_back(i);
	}
	printf ("%d %d\n",N-1,N);

	return 0;
}
### 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 <stdio.h>
#include <vector>
using namespace std;

int N; char S[100100];

int main()
{
	scanf ("%s",S+1);
	while (S[N+1]) N++;

	if (S[N] == '1' || S[1] == '0'){
		puts("-1");
		return 0;
	}

	for (int i=1;i<N;i++) if (S[i] != S[N-i]){
		puts("-1");
		return 0;
	}

	vector<int> last;
	last.push_back(1);
	for (int i=2;i<=N;i++){
		if (S[i] == '1'){
			for (auto &x : last) printf ("%d %d\n",x,i);
			last.clear();
		}
		last.push_back(i);
	}
	printf ("%d %d\n",N-1,N);

	return 0;
}
```