Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

1111

Sample Output 1

-1

It is impossible to have a connected component of size n after removing one edge from a tree with n vertices.

Sample Input 2

1110

Sample Output 2

1 2
2 3
3 4

If Edge 1 or Edge 3 is removed, we will have a connected component of size 1 and another of size 3. If Edge 2 is removed, we will have two connected components, each of size 2.

Sample Input 3

1010

Sample Output 3

1 2
1 3
1 4

Removing any edge will result in a connected component of size 1 and another of size 3.

Solution in Pascal (FPC 2.6.2) (judge verdict: Accepted)

var
    s:ansistring;
    len,i,cnt,tot,t:longint;
begin
    readln(s);len:=length(s);
    if (s[1]='0') or (s[len]='1') then
        begin
            writeln(-1);
            exit;
        end;
    dec(len);
    for i:=1 to len div 2 do
        if s[i]<>s[len-i+1] then 
            begin
                writeln(-1);
                exit;
            end;
    for i:=1 to len do
        if s[i]='1' then inc(cnt);
    for i:=1 to cnt do
        writeln(i,' ',i+1);
    tot:=cnt+2;t:=0;
    for i:=1 to len do
        if s[i]='0' then 
            begin 
                writeln(cnt-i+t+2,' ',tot);
                inc(t);inc(tot);
            end;
end.