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 (1.42.0) (judge verdict: Wrong Answer)

fn main() {
    let n = {
        let mut buf = String::new();
        std::io::stdin().read_line(&mut buf).unwrap();
        let n: usize = buf.trim_end().parse().unwrap();
        n
    };

    let s = {
        let mut buf = String::new();
        std::io::stdin().read_line(&mut buf).unwrap();
        let s = buf.trim_end().to_owned();
        s
    };

    for i in 0..1 << 2 {
        let mut l = vec![""; n];
        if i & 1 == 1 {
            l[0] = "S"
        } else {
            l[0] = "W"
        }
        if i >> 1 & 1 == 1 {
            l[1] = "S"
        } else {
            l[1] = "W"
        }
        for i in 1..n - 1 {
            let mut _c = s.chars().map(|x| x).collect::<Vec<char>>()[i];
            if l[i] == "W" {
                if _c == 'o' {
                    _c = 'x';
                } else {
                    _c = 'o';
                }
            }
            if _c == 'o' {
                l[i + 1] = l[i - 1]
            } else {
                if l[i - 1] == "S" {
                    l[i + 1] = "W"
                } else {
                    l[i + 1] = "S"
                }
            }
        }
        // 0のチェック
        let mut _c = s.chars().map(|x| x).collect::<Vec<char>>()[n - 1];
        if l[n - 1] == "W" {
            if _c == 'o' {
                _c = 'x';
            } else {
                _c = 'o';
            }
        }
        if (_c == 'o' && l[n - 2] == l[0]) || (_c == 'x' && l[n - 2] != l[0]) {
            // success
            println!("{}", l.iter().fold(String::new(), |acc, c| acc + c));
            return;
        }
        continue;
    }
    println!("{}", -1);
}