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
import java.io.*;
import java.util.*;

public class Main {
	static final Reader in = new Reader();
	static final PrintWriter out = new PrintWriter(System.out);
	
	public static void main(String[] args) {
		char[] s=in.next().toCharArray();
		int n=s.length;
		if(s[0]=='0'||s[n-1]=='1')
			fk();
		for(int i=0; i<n-2-i; ++i)
			if(s[i]!=s[n-2-i])
				fk();
		for(int i=1, l=0; i<n; ++i) {
			if(s[i]=='0')
				continue;
			for(int j=i; j>l; --j)
				out.println(j+" "+(i+1));
			l=i;
		}
		out.println(n+" "+(n-1));
		out.close();
	}
	
	static void fk() {
		System.out.println(-1);
		System.exit(0);
	}
	
	static class Reader {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st;
		String next() {
			while(st==null||!st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch(Exception e) {}
			}
			return st.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
	}
}