#include<bits/stdc++.h>
#define ll long long
using namespace std;
const ll MOD = 1e9+7;

int main(){
  string s;
  cin>>s;
  int f=1;
  if(s[s.size()-1]=='1'||s[0]=='0')f=0;
  for(int i=0;i<s.size()-1;i++){
	if(s[i]!=s[s.size()-2-i])f=0;
  }
  if(f==0){
	cout<<-1<<endl;
	return 0;
  }
  cout<<"1 2"<<endl;
  int k=2;
  for(int i=1;i<s.size()-1;i++){
	cout<<k<<" "<<2+i<<endl;
	if(s[i]=='1')k=2+i;
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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>
#define ll long long
using namespace std;
const ll MOD = 1e9+7;

int main(){
  string s;
  cin>>s;
  int f=1;
  if(s[s.size()-1]=='1'||s[0]=='0')f=0;
  for(int i=0;i<s.size()-1;i++){
	if(s[i]!=s[s.size()-2-i])f=0;
  }
  if(f==0){
	cout<<-1<<endl;
	return 0;
  }
  cout<<"1 2"<<endl;
  int k=2;
  for(int i=1;i<s.size()-1;i++){
	cout<<k<<" "<<2+i<<endl;
	if(s[i]=='1')k=2+i;
  }
  return 0;
}

```