#include <bits/stdc++.h>

const int MAXN = 1e5 + 10;

int n;
char s[MAXN];

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

Your task is to create 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>

const int MAXN = 1e5 + 10;

int n;
char s[MAXN];

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