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: Wrong Answer)

use std::io::{self, Stdin};
use std::str::{self, FromStr};
use std::error::Error;
use std::thread;

fn exec() {
    let mut sc = Scanner::new();
    let n: usize = sc.ne();
    let s = sc.ne::<String>().into_bytes();
    let mut ans = vec![0; n];
    fn yes(a: &Vec<u8>) {
        for &x in a {
            print!("{}", x as char);
        }
        println!("");
        std::process::exit(0);
    }
    for k in 0..4 {
        ans[0] = match k & 1 {
            0 => b'S',
            _ => b'W',
        };
        ans[n - 1] = match k & 2 {
            0 => b'S',
            _ => b'W',
        };
        let mut ok = true;
        for i in 0..n {
            let l = (i + n - 1) % n;
            let r = (i + 1) % n;
            let same = match s[i] {
                b'o' => ans[i] == b'S',
                _ => ans[i] == b'W',
            };
            let rin = match same {
                true => ans[l],
                _ => if ans[l] == b'S' { b'W' } else { b'S' },
            };
            if ans[r] == 0 {
                ans[r] = rin;
            } else {
                if ans[r] != rin {
                    ok = false;
                    break;
                }
            }
        }
        if ok {
            yes(&ans);
        };
    }
    println!("-1");
}
fn main() {
    const DEFAULT_STACK: usize = 16 * 1024 * 1024;
    let builder = thread::Builder::new();
    let th = builder.stack_size(DEFAULT_STACK);
    let handle = th.spawn(|| { exec(); }).unwrap();
    let _ = handle.join();
}

#[allow(dead_code)]
struct Scanner {
    stdin: Stdin,
    id: usize,
    buf: Vec<u8>,
}

#[allow(dead_code)]
impl Scanner {
    fn new() -> Scanner {
        Scanner {
            stdin: io::stdin(),
            id: 0,
            buf: Vec::new(),
        }
    }
    fn next_line(&mut self) -> Option<String> {
        let mut res = String::new();
        match self.stdin.read_line(&mut res) {
            Ok(0) => return None,
            Ok(_) => Some(res),
            Err(why) => panic!("error in read_line: {}", why.description()),
        }
    }
    fn next<T: FromStr>(&mut self) -> Option<T> {
        while self.buf.len() == 0 {
            self.buf = match self.next_line() {
                Some(r) => {
                    self.id = 0;
                    r.trim().as_bytes().to_owned()
                }
                None => return None,
            };
        }
        let l = self.id;
        assert!(self.buf[l] != b' ');
        let n = self.buf.len();
        let mut r = l;
        while r < n && self.buf[r] != b' ' {
            r += 1;
        }
        let res = match str::from_utf8(&self.buf[l..r]).ok().unwrap().parse::<T>() {
            Ok(s) => Some(s),
            Err(_) => {
                panic!("parse error, {:?}",
                       String::from_utf8(self.buf[l..r].to_owned()))
            }
        };
        while r < n && self.buf[r] == b' ' {
            r += 1;
        }
        if r == n {
            self.buf.clear();
        } else {
            self.id = r;
        }
        res
    }
    fn ne<T: FromStr>(&mut self) -> T {
        self.next::<T>().unwrap()
    }
}