#include<cstdio>
#include<cstring>
char a[100005];
int len;
int main() {
	scanf("%s", a);
	len = strlen(a);
	int i;
	if (a[0] == '0' || a[len - 1] == '1') {
		printf("-1");
		return 0;
	}
	for (i = 0; i < len/2; i++) {
		if (a[i] != a[len - 2 - i]) {
			printf("-1");
			return 0;
		}
	}
	int prev=1,x = 1;
	for (i = 1; i < len; i++) {
		x++;
		printf("%d %d\n", prev, x);
		if (i>1 && a[i - 1] == '1') prev = x;
	}
	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 a[100005];
int len;
int main() {
	scanf("%s", a);
	len = strlen(a);
	int i;
	if (a[0] == '0' || a[len - 1] == '1') {
		printf("-1");
		return 0;
	}
	for (i = 0; i < len/2; i++) {
		if (a[i] != a[len - 2 - i]) {
			printf("-1");
			return 0;
		}
	}
	int prev=1,x = 1;
	for (i = 1; i < len; i++) {
		x++;
		printf("%d %d\n", prev, x);
		if (i>1 && a[i - 1] == '1') prev = x;
	}
	return 0;
}
```