Problem:
Score : 500 points

Problem Statement

Snuke, who loves animals, built a zoo.

There are N animals in this zoo. They are conveniently numbered 1 through N, and arranged in a circle.
The animal numbered i (2≤i≤N-1) is adjacent to the animals numbered i-1 and i+1. Also, the animal numbered 1 is adjacent to the animals numbered 2 and N, and the animal numbered N is adjacent to the animals numbered N-1 and 1.

There are two kinds of animals in this zoo: honest sheep that only speak the truth, and lying wolves that only tell lies.

Snuke cannot tell the difference between these two species, and asked each animal the following question: "Are your neighbors of the same species?" The animal numbered i answered s_i. Here, if s_i is o, the animal said that the two neighboring animals are of the same species, and if s_i is x, the animal said that the two neighboring animals are of different species.

More formally, a sheep answered o if the two neighboring animals are both sheep or both wolves, and answered x otherwise.
Similarly, a wolf answered x if the two neighboring animals are both sheep or both wolves, and answered o otherwise.

Snuke is wondering whether there is a valid assignment of species to the animals that is consistent with these responses. If there is such an assignment, show one such assignment. Otherwise, print -1.

Constraints

3 ≤ N ≤ 10^{5}

s is a string of length N consisting of o and x.

Input

The input is given from Standard Input in the following format:

N
s

Output

If there does not exist an valid assignment that is consistent with s, print -1.
Otherwise, print an string t in the following format. The output is considered correct if the assignment described by t is consistent with s.

t is a string of length N consisting of S and W.

If t_i is S, it indicates that the animal numbered i is a sheep. If t_i is W, it indicates that the animal numbered i is a wolf.

Sample Input 1

6
ooxoox

Sample Output 1

SSSWWS

For example, if the animals numbered 1, 2, 3, 4, 5 and 6 are respectively a sheep, sheep, sheep, wolf, wolf, and sheep, it is consistent with their responses. Besides, there is another valid assignment of species: a wolf, sheep, wolf, sheep, wolf and wolf.

Let us remind you: if the neiboring animals are of the same species, a sheep answers o and a wolf answers x. If the neiboring animals are of different species, a sheep answers x and a wolf answers o.

Sample Input 2

3
oox

Sample Output 2

-1

Print -1 if there is no valid assignment of species.

Sample Input 3

10
oxooxoxoox

Sample Output 3

SSWWSSSWWS

Solution in Rust (1.15.1) (judge verdict: Accepted)

use std::str::FromStr;

trait Readable {
    fn read_item(s: &str) -> Self;
    fn read_line(s: String) -> Self where Self: std::marker::Sized {
        Self::read_item(&s)
    }
}

impl Readable for usize {
    fn read_item(s: &str) -> usize { usize::from_str(s).unwrap() }
}

impl Readable for u64 {
    fn read_item(s: &str) -> u64 { u64::from_str(s).unwrap() }
}

impl Readable for String {
    fn read_item(s: &str) -> String { s.to_string() }
    fn read_line(s: String) -> String { s }
}

impl<T1, T2> Readable for (T1, T2) where T1: Readable, T2: Readable {
    fn read_item(s: &str) -> (T1, T2) {
        let items: Vec<&str> = s.split(' ').collect();
        assert!(items.len() == 2);
        (T1::read_item(items[0]), T2::read_item(items[1]))
    }
}

fn read<T>() -> T where T: Readable {
    let mut line = String::new();
    std::io::stdin().read_line(&mut line).unwrap();

    let last = line.pop();
    match last {
        Some(ch) if ch != '\n' => line.push(ch),
        _ => ()
    }

    T::read_line(line)
}

fn verify(animals: &[u8], parities: &[u8]) -> bool {
    let last = animals.len() - 1;
    (parities[0] + animals[last] + animals[0] + animals[1]) % 2 == 0 &&
        (parities[last] + animals[last-1] + animals[last] + animals[0]) % 2 == 0
}

fn main() {
    let n: usize = read();
    let line: String = read();
    let parities: Vec<u8> = line.chars().map(|ch| {
        if ch == 'o' { 0 } else { 1 }
    }).collect();

    let mut animalss = [vec![0,0], vec![0,1], vec![1,0], vec![1,1]];
    for animals in animalss.iter_mut() {
        for i in 1..n-1 {
            let p = (animals[i-1] + animals[i] + parities[i]) % 2;
            animals.push(p);
        }
        if verify(&animals, &parities) {
            let ans: String = animals.iter()
                .map(|&p| if p == 0 { 'S' } else { 'W' })
                .into_iter().collect();
            println!("{}", ans);
            std::process::exit(0);
        }
    }

    println!("-1");
}