#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
using namespace std;
const int maxn = 100000 + 10;

int n;
char s[maxn];

inline void fail() {
  puts("-1");
  exit(0);
}

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

Develop a solution in Cpp to the problem described below:

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<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
using namespace std;
const int maxn = 100000 + 10;

int n;
char s[maxn];

inline void fail() {
  puts("-1");
  exit(0);
}

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