#include <bits/stdc++.h>

using namespace std;

int n;
char x[100005];
queue<int> q;
bool g;

int main() {
	scanf("%s", x + 1);
	n = strlen(x + 1);
	if (x[n] == '1')
		g = 1;
	for (int i = 1; i < n; i++)
		if (x[i] != x[n - i])
			g = 1;
	if (x[1] == '0')
		g = 1;
	
	if (g)
		printf("-1\n");
	else {
		for (int i = 1; i <= n; i++)
			q.push(i);
		for (int i = 2; i <= n; i++)
			if ((i <= n - i && x[i] == '1') || i == n)
				while (q.front() < i) {
					printf("%d %d\n", q.front(), i);
					q.pop();
				}
	}
	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 <bits/stdc++.h>

using namespace std;

int n;
char x[100005];
queue<int> q;
bool g;

int main() {
	scanf("%s", x + 1);
	n = strlen(x + 1);
	if (x[n] == '1')
		g = 1;
	for (int i = 1; i < n; i++)
		if (x[i] != x[n - i])
			g = 1;
	if (x[1] == '0')
		g = 1;
	
	if (g)
		printf("-1\n");
	else {
		for (int i = 1; i <= n; i++)
			q.push(i);
		for (int i = 2; i <= n; i++)
			if ((i <= n - i && x[i] == '1') || i == n)
				while (q.front() < i) {
					printf("%d %d\n", q.front(), i);
					q.pop();
				}
	}
	return 0;
}
```