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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i, z, n) for(int i = z; i < n; i++)
#define all(v) v.begin(), v.end()
typedef long long ll;
//const int INF = 1<<29;
//const int MODINF = 1000000007;
using namespace std;

int main(){
    int N;
    cin >> N;
    vector<char> s(N);
    vector<char> t_f(N + 1);
    rep(i, 0, N) cin >> s.at(i);
    vector<pair<char, char>> p(4);
    p.at(0) = {'S', 'S'};
    p.at(1) = {'S', 'W'};
    p.at(2) = {'W', 'S'};
    p.at(3) = {'W', 'W'};
    
    for (int k = 0; k < 4; k++){
        vector<char> t(N + 1);
        t.at(0) = p.at(k).first;
        char A;
        rep(i, 0, N){
            if (i == 0){
                A = p.at(k).second;
            }
            else{
                A = t.at(i - 1);
            }
            char z;
            if (t.at(i) == 'W'){
                if (s.at(i) == 'o') z = 'x';
                else z = 'o';
            }
            else{
                if (s.at(i) == 'o') z = 'o';
                else z = 'x';
            }
            if (z == 'o'){
                if (A == 'S') t.at(i + 1) = 'S';
                if (A == 'W') t.at(i + 1) = 'W';
            }
            else{
                if (A == 'S') t.at(i + 1) = 'W';
                if (A == 'W') t.at(i + 1) = 'S';
            }
        }
        if ((p.at(k).second == t.at(N - 1)) && (t.at(0) == t.at(N))){
            rep(i,0,N){
                t_f.at(i) = t.at(i);
            }
            goto SKIP;
        }
        else{
            if (k == 3){
                cout << -1 << endl;
                return 0;
            }
        }
    }
    SKIP:
    rep(i, 0, N){
        cout << t_f.at(i);
    }
    cout << endl;
}