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: Wrong Answer)

#include <algorithm>
#include <cstdio>
#include <iostream>
#include <map>
#include <cmath>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
#include <stdlib.h>
#include <stdio.h>
using namespace std;

#define pb         push_back
#define mp         make_pair
#define ll         long long
#define PI         acos(-1.0)
#define ALL(A)     ((A).begin(), (A).end())
#define vsort(v)   sort(v.begin(),v.end())
#define FOR(I,A,B) for(int I = (A); I < (B); ++I)

 
#define ll long long
#define MP make_pair

int main(void){
    int N;
    cin >> N;
    cin.ignore();
    string s;
    getline(cin, s);

    int who[N];
    bool check;
    FOR(i, 0, 4){
        if(i==0){
            who[0] = 0;
            who[1] = 0;
        }
        if(i==1){
            who[0] = 0;
            who[1] = 1;
        }
        if(i==2){
            who[0] = 1;
            who[1] = 1;
        }
        if(i==3){
            who[0] = 1;
            who[1] = 0;
        }
        FOR(j, 1, N-1){
            if(!who[j] && s[j]=='o') who[j+1] = who[j-1];
            if(!who[j] && s[j]=='x') who[j+1] = (who[j-1]+1) % 2;
            if(who[j] && s[j]=='o') who[j+1] = (who[j-1]+1) % 2;
            if(who[j] && s[j]=='x') who[j+1] = who[j-1];
        }
        check = false;
        if(!who[N-1] && s[N-1]=='o' && (who[0]==who[N-2])) check = true;
        if(!who[N-1] && s[N-1]=='x' && (who[0]!=who[N-2])) check = true;
        if(who[N-1] && s[N-1]=='o' && (who[0]!=who[N-2])) check = true;
        if(who[N-1] && s[N-1]=='x' && (who[0]==who[N-2])) check = true;
        if(!who[0] && s[0]=='o' && (who[1]==who[N-1])) check = true;
        if(!who[0] && s[0]=='x' && (who[1]!=who[N-1])) check = true;
        if(who[0] && s[0]=='o' && (who[1]!=who[N-1])) check = true;
        if(who[0] && s[0]=='x' && (who[1]==who[N-1])) check = true;
        if(check) break;
    }
    if(!check){
        cout << "-1" << endl;
        return 0;
    }
    FOR(i, 0, N) cout << (who[i] ? "W" : "S");
    cout << endl;
}