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)

#[allow(unused_imports)]
use std::cmp::{max, min};
#[allow(unused_imports)]
use std::collections::{HashMap, HashSet};

mod util {
    use std::io::stdin;
    use std::str::FromStr;
    use std::fmt::Debug;

    #[allow(dead_code)]
    pub fn line() -> String {
        let mut line: String = String::new();
        stdin().read_line(&mut line).unwrap();
        line.trim().to_string()
    }

    #[allow(dead_code)]
    pub fn get<T: FromStr>() -> T
        where <T as FromStr>::Err: Debug
    {
        let mut line: String = String::new();
        stdin().read_line(&mut line).unwrap();
        line.trim().parse().unwrap()
    }

    #[allow(dead_code)]
    pub fn gets<T: FromStr>() -> Vec<T>
        where <T as FromStr>::Err: Debug
    {
        let mut line: String = String::new();
        stdin().read_line(&mut line).unwrap();
        line.split_whitespace()
            .map(|t| t.parse().unwrap())
            .collect()
    }
}

fn main() {
    let _: usize = util::get();
    let s: Vec<bool> = util::line().chars().map(|c| c == 'o').collect();

    let ans = [vec![true, true],
               vec![true, false],
               vec![false, true],
               vec![false, false]]
            .into_iter()
            .cloned()
            .map(|mut seq| {
                     for &x in &s {
                         let (a, b) = (seq[seq.len() - 2], seq[seq.len() - 1]);
                         let c = a ^ b ^ x;
                         seq.push(c);
                     }
                     seq
                 })
            .find(|seq| seq[0] == seq[seq.len() - 2] && seq[1] == seq[seq.len() - 1])
            .map(|mut seq| {
                     seq.remove(0);
                     seq.pop();
                     seq
                 });

    if let Some(seq) = ans {
        println!("{}",
                 seq.into_iter()
                     .map(|b| if b { 'S' } else { 'W' })
                     .collect::<String>());
    } else {
        println!("-1");
    }
}