#include<bits/stdc++.h>
using namespace std;
int main(){
  int n,j,k;
  string s;
  cin>>n>>s;
  vector<int> x(n+1);
  for(j=1;j<=n;j++){
    if(s.at(j-1)=='o') x.at(j)=1;
    else if(s.at(j-1)=='x') x.at(j)=-1;
  }
  vector<int> s1_={1,1,-1,-1};
  vector<int> s2_={1,-1,1,-1};
  for(k=0;k<4;k++){
    vector<int> y(n+1);
    y.at(1)=s1_.at(k);
    y.at(2)=s2_.at(k);
    y.at(0)=y.at(2)*y.at(1)*x.at(1);
    for(j=3;j<=n;j++) y.at(j)=y.at(j-2)*y.at(j-1)*x.at(j-1);
    if(y.at(0)==y.at(n)&&y.at(1)==y.at(n-1)*y.at(n)*x.at(n)){
      for(j=1;j<=n;j++){
        if(y.at(j)==1) cout<<'S';
        else if(y.at(j)==-1) cout<<'W';
      }
      cout<<endl;
      return 0;
    }
  }
  cout<<-1<<endl;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
int main(){
  int n,j,k;
  string s;
  cin>>n>>s;
  vector<int> x(n+1);
  for(j=1;j<=n;j++){
    if(s.at(j-1)=='o') x.at(j)=1;
    else if(s.at(j-1)=='x') x.at(j)=-1;
  }
  vector<int> s1_={1,1,-1,-1};
  vector<int> s2_={1,-1,1,-1};
  for(k=0;k<4;k++){
    vector<int> y(n+1);
    y.at(1)=s1_.at(k);
    y.at(2)=s2_.at(k);
    y.at(0)=y.at(2)*y.at(1)*x.at(1);
    for(j=3;j<=n;j++) y.at(j)=y.at(j-2)*y.at(j-1)*x.at(j-1);
    if(y.at(0)==y.at(n)&&y.at(1)==y.at(n-1)*y.at(n)*x.at(n)){
      for(j=1;j<=n;j++){
        if(y.at(j)==1) cout<<'S';
        else if(y.at(j)==-1) cout<<'W';
      }
      cout<<endl;
      return 0;
    }
  }
  cout<<-1<<endl;
}
```