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)

#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
using namespace std;
using ll=long long;
using datas=pair<ll,ll>;
using ddatas=pair<double,double>;
using tdata=pair<ll,datas>;
using vec=vector<ll>;
using mat=vector<vec>;
using pvec=vector<datas>;
using pmat=vector<pvec>;
#define For(i,a,b) for(i=a;i<b;i++)
#define bFor(i,a,b) for(i=a;i>=b;i--)
#define rep(i,N) For(i,0,N)
#define rep1(i,N) For(i,1,N)
#define brep(i,N) bFor(i,N-1,0)
#define all(v) (v).begin(),(v).end()
#define allr(v) (v).rbegin(),(v).rend()
#define vsort(v) sort(all(v))
#define vrsort(v) sort(allr(v))
#define endl "\n"
#define pb push_back
#define output(v) do{bool f=0;for(auto outi:v){cout<<(f?" ":"")<<outi;f=1;}cout<<endl;}while(0)
const ll mod=1000000007;
const ll inf=1LL<<60;
template<class T> inline bool chmax(T& a,T b){bool x=a<b;if(x)a=b;return x;} 
template<class T> inline bool chmin(T& a,T b){bool x=a>b;if(x)a=b;return x;} 

double distance(ddatas& x,ddatas& y){
  double a=x.first-y.first,b=x.second-y.second;
  return sqrt(a*a+b*b);
}

ll modinv(ll a) {
  ll b=mod,u=1,v=0,t;
  while(b){
    t=a/b;
    a-=t*b; swap(a,b);
    u-=t*v; swap(u,v);
  }
  return (u+mod)%mod;
}

ll moddevide(ll a,ll b){return (a*modinv(b))%mod;}

vec modncrlistp,modncrlistm;

ll modncr(ll n,ll r){
  ll i,size=modncrlistp.size();
  if(size<=n){
    modncrlistp.resize(n+1);
    modncrlistm.resize(n+1);
    if(!size){
      modncrlistp[0]=modncrlistm[0]=1;
      size++;
    }
    For(i,size,n+1){
      modncrlistp[i]=modncrlistp[i-1]*i%mod;
      modncrlistm[i]=modinv(modncrlistp[i]);
    }
  }
  return modncrlistp[n]*modncrlistm[r]%mod*modncrlistm[n-r]%mod;
}

ll modpow(ll a,ll n){
  ll res=1;
  while(n){
    if(n&1)res=res*a%mod;
    a=a*a%mod;
    n>>=1;
  }
  return res;
}

ll gcd(ll a,ll b){if(!b)return a;return (a%b==0)?b:gcd(b,a%b);}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}

ll countdigits(ll n){
  ll ans=0;
  while(n){n/=10;ans++;}
  return ans;
}

ll sumdigits(ll n){
  ll ans=0;
  while(n){ans+=n%10;n/=10;}
  return ans;
}

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  ll i,j,N,M=4;
  cin>>N;
  string s;
  cin>>s;
  map<ll,ll> v;
  rep(j,M){
    v[-1]=j%2;
    v[0]=j/2%2;
    rep1(i,N+1){
      if(s[i-1]=='o'){
        if(v[i-1]){
          v[i]=v[i-2];
        }else{
          v[i]=v[i-2]^1;
        }
      }else{
        if(v[i-1]){
          v[i]=v[i-2]^1;
        }else{
          v[i]=v[i-2];
        }
      }
    }
    if(v[0]!=v[N]||v[-1]!=v[N-1])continue;
    rep(i,N){
      printf("%s",(v[i])?"S":"W");
    }
    cout<<endl;
    return 0;
  }
  cout<<-1<<endl;
  return 0;
}