#include <bits/stdc++.h>
using namespace std;
 
int main(){
  string s; cin >> s;
  int n = s.length();
  if(s[n-1]=='1' || s[n-2]=='0' || s[0] =='0'){ cout << "-1" << endl; return 0;}
  for(int i=0; i<n/2;i++){
  	if(s[n-2-i]!=s[i]) {cout << "-1" << endl; return 0;}
  }
  
  s[n-1] = '1';
  vector<int> v;
  for(int i=0;i<n;i++) if(s[i]=='1') v.push_back(i);
  for(int i=0;i<n-1;i++){
    int idx = upper_bound(v.begin(), v.end(), i) - v.begin();
  	cout << i+1 << " " << v[idx]+1 << endl;
  }
  return 0;
}
### Prompt

In Cpp, your task is to solve 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;
 
int main(){
  string s; cin >> s;
  int n = s.length();
  if(s[n-1]=='1' || s[n-2]=='0' || s[0] =='0'){ cout << "-1" << endl; return 0;}
  for(int i=0; i<n/2;i++){
  	if(s[n-2-i]!=s[i]) {cout << "-1" << endl; return 0;}
  }
  
  s[n-1] = '1';
  vector<int> v;
  for(int i=0;i<n;i++) if(s[i]=='1') v.push_back(i);
  for(int i=0;i<n-1;i++){
    int idx = upper_bound(v.begin(), v.end(), i) - v.begin();
  	cout << i+1 << " " << v[idx]+1 << endl;
  }
  return 0;
}
```