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::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 _ = scan.read::<usize>();
    let s = scan.chars();
    let c = vec![-1, 1];
    let mut ret = None;
    for &i in &c {
        for &j in &c {
            if let Some(ans) = check(i, j, &s) {
                ret = Some(ans);
                break;
            }
        }
    }
    if let Some(ans) = ret {
        println!("{}", ans.iter().map(|&x| if x == -1 {'S'} else {'W'}).collect::<String>());
    } else {
        println!("{}", -1);
    }
}

fn check(a0: i32, a1: i32, s: &Vec<char>) -> Option<Vec<i32>> {
    let n = s.len();
    let mut ans = vec![0; n];
    ans[0] = a0;
    ans[1] = a1;
    for i in 1..n-1 {
        match (s[i], ans[i]) {
            ('o', -1) => ans[i+1] = ans[i-1],
            ('x', -1) => ans[i+1] = -ans[i-1],
            ('o', 1) => ans[i+1] = -ans[i-1],
            _ => ans[i+1] = ans[i-1]
        }
    }
    for i in n-1..n+1 {
        if !(match (s[i%n], ans[i%n]) {
            ('o', -1) => {
                ans[(i+1)%n] == ans[i-1]
            },
            ('x', -1) => {
                ans[(i+1)%n] != ans[i-1]
            },
            ('o', 1) => {
                ans[(i+1)%n] != ans[i-1]
            },
            _ => {
                ans[(i+1)%n] == ans[i-1]
            }
        }) {return None;}
    }
    Some(ans)
}

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

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

#[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