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.

Examples

Input

6
ooxoox


Output

SSSWWS


Input

3
oox


Output

-1


Input

10
oxooxoxoox


Output

SSWWSSSWWS
#include <iostream>
#include <cmath>
#include <vector>
#include <map>
#include <stack>
#include <queue>
#include <set>
#include <algorithm>
#include <iomanip>
#include <string.h>

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define ALL(a)  (a).begin(),(a).end()

typedef long long lint;

using namespace std;

int main(){
  int N;
  string s;
  cin>>N;
  cin>>s;
  string t[4];
  t[0]="SS";
  t[1]="SW";
  t[2]="WS";
  t[3]="WW";
  FOR(i,1,N){
    REP(j,4){
      if((s[i]=='o')^(t[j][i]=='S')) t[j]+=(t[j][i-1]=='S'?'W':'S');
      else  t[j]+=(t[j][i-1]=='S'?'S':'W');
    }
  }
  // REP(i,4)cerr<<t[i]<<endl;
  REP(i,4){
    bool f=(t[i][0]==t[i][N])and(((s[0]=='o')^(t[i][0]=='S'))==(t[i][1]!=t[i][N-1]));
    if(!f)continue;
    cout<<t[i].substr(0,N)<<endl;
    return 0;
  }
  cout<<"-1"<<endl;
  return 0;
}
