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

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
 
import static java.lang.Math.*;
import static java.util.Arrays.*;
import static java.util.Collections.*;
 
public class Main{
    static final long mod=1000000007;
    
    public  static void main(String[] args)   throws Exception, IOException{
        Reader sc = new Reader(System.in);
        PrintWriter out=new PrintWriter(System.out);
        // int n=sc.nextInt();
        // char c[][] = new char[h][w];
        // char c[]=sc.nextString().toCharArray();
        // int d[]=new int[n];
        // for(int i=0; i<n; i++) {d[i]=sc.nextInt();}

        // int n=sc.nextInt();
        char c[]=sc.nextString().toCharArray();
        int l=c.length;

        if(c[0]=='0' || c[l-1]=='1'){out.println(-1);out.flush();return;}
        else {
            for (int i=0; i<l/2; i++) {
                if(c[i]!=c[l-i-2]){out.println(-1);out.flush(); return;}
            }
        }

        out.println(1+" "+2);

        for (int i=3,s=2; i<=l; i++) {
            out.println(s+" "+i);
            if(c[i-2]=='1'){
                s=i;
            }
        }


      
        // out.println(ans);
        out.flush();
    }


    static void db(Object... os){
        System.err.println(Arrays.deepToString(os));
    }  

    static long gcd(long a, long b){
        if(min(a,b) == 0)return max(a,b);
        return gcd(max(a,b) % min(a,b),min(a,b));
    }
}

class P implements Comparable<P>{
    int id, d;
    P(int  id, int d) {
        this.id=id;
        this.d=d;
    }
    public int compareTo(P p){
        return d-p.d; //des
    }
}

class Reader
{ 
    private BufferedReader x;
    private StringTokenizer st;
    
    public Reader(InputStream in)
    {
        x = new BufferedReader(new InputStreamReader(in));
        st = null;
    }
    public String nextString() throws IOException
    {
        while( st==null || !st.hasMoreTokens() )
            st = new StringTokenizer(x.readLine());
        return st.nextToken();
    }
    public int nextInt() throws IOException
    {
        return Integer.parseInt(nextString());
    }
    public long nextLong() throws IOException
    {
        return Long.parseLong(nextString());
    }
    public double nextDouble() throws IOException
    {
        return Double.parseDouble(nextString());
    }
}