#include <stdio.h>
#include <cstring>

const int MAXN = 100002;

char s[MAXN];

int main ()
{

	scanf("%s", s);
	int n = strlen(s);

	if(s[n - 1] == '1' || s[0] == '0') {
		printf("-1");
		return 0;
	}
	bool rau = false;
	for(int i = 0; i < n / 2; i++) {
		if(s[i] != s[n - 2 - i]) {
			rau = true;
		}
	}
	if(rau) {
		printf("-1");
		return 0;
	}
	printf("1 2\n");
	int nod = 2;
	for(int i = 1; i < n - 1; i++) {
		printf("%d %d\n", i + 2, nod);
		if(s[i] == '1') {
			nod = i + 2;
		}
	}

	// 7 1011010
	// 8 110101100.
	return 0;
}
### Prompt

Please provide a CPP coded solution 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 <stdio.h>
#include <cstring>

const int MAXN = 100002;

char s[MAXN];

int main ()
{

	scanf("%s", s);
	int n = strlen(s);

	if(s[n - 1] == '1' || s[0] == '0') {
		printf("-1");
		return 0;
	}
	bool rau = false;
	for(int i = 0; i < n / 2; i++) {
		if(s[i] != s[n - 2 - i]) {
			rau = true;
		}
	}
	if(rau) {
		printf("-1");
		return 0;
	}
	printf("1 2\n");
	int nod = 2;
	for(int i = 1; i < n - 1; i++) {
		printf("%d %d\n", i + 2, nod);
		if(s[i] == '1') {
			nod = i + 2;
		}
	}

	// 7 1011010
	// 8 110101100.
	return 0;
}
```