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 (1.15.1) (judge verdict: Accepted)

use std::io::{stdin, Read, StdinLock};
use std::str::FromStr;
use std::iter::FromIterator;

fn main() {
    let cin = stdin();
    let mut scan = Scanner::new(cin.lock());
    let s = scan.chars();
    let n = s.len();
    let mut is_ok = true;
    if s[0] != '1' || s[n-1] != '0' {
        is_ok = false;
    }
    for i in 0..n {
        if i >= n-2-i {break;}
        if s[i] != s[n-2-i] {
            is_ok = false;
        }
    }
    if !is_ok {
        println!("-1");
        return;
    }
    let mut leaf = vec![];
    let mut edge = vec![];
    for i in 0..n {
        let mut c = s[i];
        if i == n-1 {c = '1';}
        if c == '0' {
            leaf.push(i);
        } else {
            for &v in &leaf {
                edge.push((v, i));
            }
            leaf = vec![i];
        }
    }
    for &(p, q) in &edge {
        println!("{} {}", p+1, q+1);
    }
}

// region template
#[allow(dead_code)]
fn iu(i: i64) -> usize { i as usize }

#[allow(dead_code)]
fn ui(i: usize) -> i64 { i as i64 }

#[allow(dead_code, deprecated)]
fn join<T: std::fmt::Display>(slice: &[T], sep: &str) -> String {
    let strings = slice.iter().map(|t| format!("{}", t)).collect::<Vec<_>>();
    strings.connect(sep)
}

#[allow(dead_code)]
fn arr<'a, S, T>(n: usize, mut f: S) -> Vec<T> where S: FnMut(usize) -> T + 'a {
    (0..n).map(|i| f(i)).collect::<Vec<T>>()
}

#[allow(dead_code)]
fn alt<S, T>(v: Vec<S>) -> T where T: FromIterator<S> {
    v.into_iter().collect::<T>()
}

#[allow(dead_code)]
fn dup<S, T>(v: &[S]) -> T where T: FromIterator<S>, S: Clone {
    v.iter().cloned().collect::<T>()
}

struct Scanner<'a> {
    cin: StdinLock<'a>,
}

#[allow(dead_code)]
impl<'a> Scanner<'a> {
    fn new(cin: StdinLock<'a>) -> Scanner<'a> {
        Scanner { cin: cin }
    }

    fn read1<T: FromStr>(&mut self) -> Option<T> {
        let token = self.cin.by_ref().bytes().map(|c| c.unwrap() as char)
            .skip_while(|c| c.is_whitespace())
            .take_while(|c| !c.is_whitespace())
            .collect::<String>();
        token.parse::<T>().ok()
    }

    fn read<T: FromStr>(&mut self) -> T {
        self.read1().unwrap()
    }

    fn readn<T: FromStr>(&mut self, n: usize) -> Vec<T> {
        (0..n).map(|_| self.read::<T>()).collect()
    }

    fn chars(&mut self) -> Vec<char> {
        self.read::<String>().chars().collect()
    }
}
// endregion