#include <bits/stdc++.h>

using namespace std;

const int maxn = 1e5 + 5;
char s[maxn];
int n;

bool judge() {
	for (int i = 1; i < n; ++i)
		if (s[i] != s[n-i]) return true;
	return false;
}

int main() {
	cin >> (s+1);
	for (n = 1; s[n] != '\0'; ++n);
	--n;
	if (s[n] != '0' || s[1] != '1' || judge()) puts("-1");
	else {
		int root = 1;
		printf("1 2\n");
		for (int i = 2; i < n; ++i) {
			printf("%d %d\n", root, i+1);
			if (s[i] == '1') root = i+1;
		}
	}
	return 0;
}
### Prompt

Your challenge is to write 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 <bits/stdc++.h>

using namespace std;

const int maxn = 1e5 + 5;
char s[maxn];
int n;

bool judge() {
	for (int i = 1; i < n; ++i)
		if (s[i] != s[n-i]) return true;
	return false;
}

int main() {
	cin >> (s+1);
	for (n = 1; s[n] != '\0'; ++n);
	--n;
	if (s[n] != '0' || s[1] != '1' || judge()) puts("-1");
	else {
		int root = 1;
		printf("1 2\n");
		for (int i = 2; i < n; ++i) {
			printf("%d %d\n", root, i+1);
			if (s[i] == '1') root = i+1;
		}
	}
	return 0;
}
```