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::{Ordering};

#[allow(unused_imports)]
use std::collections::{BTreeMap, BTreeSet, BinaryHeap, HashMap, HashSet, VecDeque};

#[allow(unused_imports)]
use std::io::{stdin, stdout, BufWriter, Write};

#[allow(unused_imports)]
use std::iter::FromIterator;

#[allow(unused_macros)]
macro_rules! min {
    ($x: expr) => ($x);
    ($x: expr, $($z: expr),+) => {{
        std::cmp::min($x, min!($($z),*))
    }}
}

#[allow(unused_macros)]
macro_rules! max {
    ($x: expr) => ($x);
    ($x: expr, $($z: expr),+) => {{
        std::cmp::max($x, max!($($z),*))
    }}
}

#[allow(unused_macros)]
macro_rules! get { 
    ([$t: ty]) => { 
        { 
            let mut line = String::new(); 
            stdin().read_line(&mut line).unwrap(); 
            line.split_whitespace().map(|t|t.parse::<$t>().unwrap()).collect::<Vec<_>>()
        }
    };
    ([$t: ty];$n: expr) => {
        (0..$n).map(|_|get!([$t])).collect::<Vec<_>>()
    };
    ($t: ty) => {
        {
            let mut line = String::new();
            stdin().read_line(&mut line).unwrap();
            line.trim().parse::<$t>().unwrap()
        }
    };
    ($($t: ty),*) => {
        { 
            let mut line = String::new();
            stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            ($(iter.next().unwrap().parse::<$t>().unwrap(),)*)
        }
    };
    ($t: ty; $n: expr) => {
        (0..$n).map(|_|get!($t)).collect::<Vec<_>>()
    };
    ($($t: ty),*; $n: expr) => {
        (0..$n).map(|_|get!($($t),*)).collect::<Vec<_>>()
    };
}

#[allow(unused_macros)]
macro_rules! debug {
    ($($a:expr),*) => {
        #[cfg(debug_assertions)]
        writeln!(&mut std::io::stderr(), concat!("[DEBUG] ", $(stringify!($a), "={:?} "),*), $($a),*).unwrap();
    }
}

const BIG_STACK_SIZE: bool = true;

#[allow(dead_code)]
fn main() {
    use std::thread;
    if BIG_STACK_SIZE {
        thread::Builder::new()
            .stack_size(32 * 1024 * 1024)
            .name("solve".into())
            .spawn(solve)
            .unwrap()
            .join()
            .unwrap();
    } else {
        solve();
    }
}

fn show(vec: Vec<bool>) {
    for e in vec {
        if e {
            print!("S");
        } else {
            print!("W");
        }
    }
    println!("");
}

fn solve() {
    let n = get!(usize);
    let s = get!(String).chars().collect::<Vec<_>>();

    for i in 0..4 {
        let mut vec = vec![false; n];
        vec[0] = i % 2 == 0;
        vec[1] = (i>>1) % 2 == 0;

        for i in 2..n {
            vec[i] = if s[i-1] == 'o' {
                if vec[i-1] {
                    vec[i-2]
                } else {
                    !vec[i-2]
                }
            } else {
                if vec[i-1] {
                    !vec[i-2]
                } else {
                    vec[i-2]
                }
            };
        }

        let ok0 = if s[0] == 'o' {
            if vec[0] {
                vec[1] == vec[n-1]
            } else {
                vec[1] != vec[n-1]
            }
        } else {
            if vec[0] {
                vec[1] != vec[n-1]
            } else {
                vec[1] == vec[n-1]
            }
        };

        let okn = if s[n-1] == 'o' {
            if vec[n-1] {
                vec[n-2] == vec[0]
            } else {
                vec[n-2] != vec[0]
            }
        } else {
            if vec[n-1] {
                vec[n-2] != vec[0]
            } else {
                vec[n-2] == vec[0]
            }
        };

        if ok0 && okn {
            show(vec);
            return;
        }
    }
    println!("-1");
}