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::vec::Vec;

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>().split_whitespace().map(|e| e.parse().ok().unwrap()).collect()
}

fn main() {
    let n: usize = read();
    let mut s: Vec<char> = read::<String>().chars().collect();

    // s -> true
    // w -> false
    let sw = vec![true, false];
    let mut check = vec![false; n];

    for &first in &sw {
        'label: for &second in &sw {
            check[0] = first;
            check[1] = second;
            // 前の２つにしたがって割り付ける
            for i in 2..n {
                let a = check[i-2]; // ２つ前
                let b = check[i-1]; // １つ前
                allocation(a, b, i, &s, &mut check);
            }


            // checkの割付がsと矛盾しないかを確認
            for i in 0..n {
                let (pre, now, nxt) = if i == 0 {
                   (check[n-1], check[i], check[i+1])
                } else if i == n-1 {
                   (check[i-1], check[i], check[0])
                } else {
                   (check[i-1], check[i], check[i+1])
                };
                
                match (s[i], now) {
                    ('o', true) => {
                        if nxt != pre {
                           continue 'label;
                        }
                    },
                    ('o', false) => {
                        if nxt == pre {
                           continue 'label;
                        }
                    }

                    ('x', true) => {
                        if nxt == pre {
                           continue 'label;
                        }
                    },
                    ('x', false) => {
                        if nxt != pre {
                           continue 'label;
                        }
                    }
                    _  => panic!(),
                }               
            }
            let ans: String = check.into_iter().map(|b| if b { 'S' } else { 'W' }).collect();
            println!("{}", ans);
            return
        }
    }
    println!("-1");
}

// ２つ前と１つ前から現在の要素を確定させる
fn allocation(a: bool, b: bool, index: usize, s: &Vec<char>, check: &mut Vec<bool>) {
    let pre = s[index-1];

    match (pre, b) {
        ('o', true) => check[index] = a,
        ('x', true) => check[index] = !a,
        ('o', false) => check[index] = !a,
        ('x', false) => check[index] = a,
        _ => panic!(),
    }
}