#include <bits/stdc++.h>

std::string s;

bool able();

int main()
{
	std::cin >> s;
	if (!able())
	{
		printf("-1\n");
		return 0;
	}
	int pre{0};
	for (unsigned int i = 0; i < s.size() - 1; i++)
	{
		printf("%d %d\n", pre + 1, i + 2);
		if (s[i] == '1') pre = i + 1;
	}

	return 0;
}

bool able()
{
	if (s.front() == '0' || *(s.end() - 2) == '0' || s.back() == '1') return false;
	for (unsigned int i = 0; i < s.size() - 1; i++)
		if (s[i] != s[s.size() - i - 2]) return false;
	return true;
}
### Prompt

Create a solution in Cpp for 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 <bits/stdc++.h>

std::string s;

bool able();

int main()
{
	std::cin >> s;
	if (!able())
	{
		printf("-1\n");
		return 0;
	}
	int pre{0};
	for (unsigned int i = 0; i < s.size() - 1; i++)
	{
		printf("%d %d\n", pre + 1, i + 2);
		if (s[i] == '1') pre = i + 1;
	}

	return 0;
}

bool able()
{
	if (s.front() == '0' || *(s.end() - 2) == '0' || s.back() == '1') return false;
	for (unsigned int i = 0; i < s.size() - 1; i++)
		if (s[i] != s[s.size() - i - 2]) return false;
	return true;
}
```