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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;


int main() {
    int N;
    string s;
    cin >> N >> s;

    // 1, 2, N
    vector<tuple<char,char,char>> inits;
    if (s[0] == 'o') {
        inits.push_back(make_tuple('S', 'S', 'S'));
        inits.push_back(make_tuple('S', 'W', 'W'));
        inits.push_back(make_tuple('W', 'S', 'W'));
        inits.push_back(make_tuple('W', 'W', 'S'));
    } else {
        inits.push_back(make_tuple('S', 'S', 'W'));
        inits.push_back(make_tuple('S', 'W', 'S'));
        inits.push_back(make_tuple('W', 'S', 'S'));
        inits.push_back(make_tuple('W', 'W', 'W'));
    }
    bool is_ok = false;
    vector<char> a(N);
    for (auto init : inits) {
        a = vector<char>(N);
        a[0] = get<0>(init);
        a[1] = get<1>(init);
        char a_N = get<2>(init);
        for (int i = 1; i < N - 1; i++) {
            if (a[i] == 'S') {
                if (s[i] == 'o') {
                    a[i + 1] = a[i - 1];
                } else {
                    a[i + 1] = a[i - 1] == 'S' ? 'W' : 'S';
                }
            } else { // 'W'
                if (s[i] == 'o') {
                    a[i + 1] = a[i - 1] == 'S' ? 'W' : 'S';
                } else {
                    a[i + 1] = a[i - 1];
                }
            }
        }
        if (a[N - 1] != a_N) continue;
        if (a[N - 1] == 'S') {
            if (s[N - 1] == 'o') {
                if (a[N - 2] == a[0]) { is_ok = true; break; }
            } else {
                if (a[N - 2] != a[0]) { is_ok = true; break; }
            }
        } else { // 'W'
            if (s[N - 1] == 'o') {
                if (a[N - 2] != a[0]) { is_ok = true; break; }
            } else {
                if (a[N - 2] == a[0]) { is_ok = true; break; }
            }
        }
    }
    if (is_ok) {
        for (char c : a) cout << c;
        cout << endl;
    } else {
        cout << -1 << endl;
    }

    return 0;
}