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 <bits/stdc++.h>
using namespace std;

typedef long long ll;
#define all(x) (x).begin(),(x).end()
const int mod=1000000007;

int main(){
    
    int N;cin>>N;
    char T[N];
    for(int i=0;i<N;i++){
        cin>>T[i];
    }
    string A="SS",B="SW",C="WS",D="WW";
    for(int i=1;i<N;i++){
        if(T[i]=='o'){
            if(A[i]=='S') A+=A[i-1];
            else if(A[i-1]=='S') A+='W';
            else A+='S';
            if(B[i]=='S') B+=B[i-1];
            else if(B[i-1]=='S') B+='W';
            else B+='S';
            if(C[i]=='S') C+=C[i-1];
            else if(C[i-1]=='S') C+='W';
            else C+='S';
            if(D[i]=='S') D+=D[i-1];
            else if(D[i-1]=='S') D+='W';
            else D+='S';
        }else{
            if(A[i]=='W') A+=A[i-1];
            else if(A[i-1]=='S') A+='W';
            else A+='S';
            if(B[i]=='W') B+=B[i-1];
            else if(B[i-1]=='S') B+='W';
            else B+='S';
            if(C[i]=='W') C+=C[i-1];
            else if(C[i-1]=='S') C+='W';
            else C+='S';
            if(D[i]=='W') D+=D[i-1];
            else if(D[i-1]=='S') D+='W';
            else D+='S';
        }
    }
    if(T[0]=='o'){
        if(A[1]==A[N]&&A[0]==A[N]){
            A.pop_back();
            cout<<A<<endl;
        }else if(B[1]==B[N]&&B[0]==B[N]){
            B.pop_back();
            cout<<B<<endl;
        }else if(C[1]!=C[N]&&C[0]==C[N]){
            C.pop_back();
            cout<<C<<endl;
        }else if(D[1]!=D[N]&&D[0]==D[N]){
            D.pop_back();
            cout<<D<<endl;
        }else cout<<-1<<endl;
    }else{
        if(A[1]!=A[N]&&A[0]==A[N]){
            A.pop_back();
            cout<<A<<endl;
        }else if(B[1]!=B[N]&&B[0]==B[N]){
            B.pop_back();
            cout<<B<<endl;
        }else if(C[1]==C[N]&&C[0]==C[N]){
            C.pop_back();
            cout<<C<<endl;
        }else if(D[1]==D[N]&&D[0]==D[N]){
            D.pop_back();
            cout<<D<<endl;
        }else cout<<-1<<endl;
        
    }
}