#include <iostream>
using namespace std;
typedef long long ll;
ll n,cnt=1;
string s;
int main(void){
    cin>>s;
    n=s.size();
    if(n==2&&s=="10"){
        cout<<"1 2"<<endl;
        return 0;
    }
    if(s[n-1]=='1'||s[0]=='0'){
        cout<<-1<<endl;
        return 0;
    }
    for(int i=0;i<(n-1)/2;i++){
        if(s[i]!=s[n-i-2]){
            cout<<-1<<endl;
            return 0;
        }
    }
    for(int i=0;i<n-1;i++){
        cout<<cnt<<" "<<i+2<<endl;
        if(s[i]=='1'){
            cnt=i+2;
        }
    }
}

### 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 <iostream>
using namespace std;
typedef long long ll;
ll n,cnt=1;
string s;
int main(void){
    cin>>s;
    n=s.size();
    if(n==2&&s=="10"){
        cout<<"1 2"<<endl;
        return 0;
    }
    if(s[n-1]=='1'||s[0]=='0'){
        cout<<-1<<endl;
        return 0;
    }
    for(int i=0;i<(n-1)/2;i++){
        if(s[i]!=s[n-i-2]){
            cout<<-1<<endl;
            return 0;
        }
    }
    for(int i=0;i<n-1;i++){
        cout<<cnt<<" "<<i+2<<endl;
        if(s[i]=='1'){
            cnt=i+2;
        }
    }
}

```