#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

char inp[100005];

int main() {
	scanf("%s", inp + 1);
	int N = strlen(inp + 1);
	if (inp[N] == '1' || inp[1] == '0') return !printf("-1");
	for (int i = 1; i < N; i++) {
		if (inp[i] != inp[N - i]) return !printf("-1");
	}
	int used = 1;
	for (int i = 2; i < N; i++) {
		if (inp[i] == '1') {
			while (used < i) printf("%d %d\n", used++, i);
		}
	}
	printf("%d %d\n", N - 1, N);
}
### Prompt

Generate 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 <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

char inp[100005];

int main() {
	scanf("%s", inp + 1);
	int N = strlen(inp + 1);
	if (inp[N] == '1' || inp[1] == '0') return !printf("-1");
	for (int i = 1; i < N; i++) {
		if (inp[i] != inp[N - i]) return !printf("-1");
	}
	int used = 1;
	for (int i = 2; i < N; i++) {
		if (inp[i] == '1') {
			while (used < i) printf("%d %d\n", used++, i);
		}
	}
	printf("%d %d\n", N - 1, N);
}
```