#include <bits/stdc++.h>
using namespace std;
void fuk() { puts("-1"); exit(0); }

char A[100055];

int N, C = 1;

int f(int cnt, int rt) {
	if(1 == cnt) return 1;
	int nt = cnt-1; for(; '0' == A[nt]; nt--);
	C++; printf("%d %d\n", rt, C);
	int lft = cnt - 1 - f(nt, C);
	for(; lft--;) {
		C++;
		printf("%d %d\n", rt, C);
	}
	return cnt;
}

int main() {
	scanf(" %s", A+1);
	N = strlen(A+1);

	if('1' != A[1] || '0' != A[N]) fuk();
	for(int i = 1; i < N; i++)
		if(A[i] != A[N-i]) fuk();

	f(N, 1);

	return 0;
}
### Prompt

Generate 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;
void fuk() { puts("-1"); exit(0); }

char A[100055];

int N, C = 1;

int f(int cnt, int rt) {
	if(1 == cnt) return 1;
	int nt = cnt-1; for(; '0' == A[nt]; nt--);
	C++; printf("%d %d\n", rt, C);
	int lft = cnt - 1 - f(nt, C);
	for(; lft--;) {
		C++;
		printf("%d %d\n", rt, C);
	}
	return cnt;
}

int main() {
	scanf(" %s", A+1);
	N = strlen(A+1);

	if('1' != A[1] || '0' != A[N]) fuk();
	for(int i = 1; i < N; i++)
		if(A[i] != A[N-i]) fuk();

	f(N, 1);

	return 0;
}
```