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 Rust (1.42.0) (judge verdict: Accepted)

use proconio::input;

fn err() -> ! {
    println!("{}", -1);
    std::process::exit(0);
}

fn main() {
    input!(s: String);
    let n = s.len();
    if s.as_bytes()[n - 1] == b'1' || s.as_bytes()[0] == b'0' {
        err();
    }
    if s.chars().zip(s.chars().rev().skip(1)).any(|(x, y)| x != y) {
        err();
    }
    let mut next = vec![n - 1; n];
    for i in (0..n - 1).rev() {
        next[i] = if s.as_bytes()[i + 1] == b'1' {
            i + 1
        } else {
            next[i + 1]
        };
        println!("{} {}", i + 1, next[i] + 1);
    }
}