#include <iostream>
#include <vector>
using namespace std;

int main(){
    int N;
    string S;
    cin >> N >> S;
    vector<bool> responce;
    for(int i = 0; i < N; i++){
        responce.push_back(S[i] == 'o');
    }
    for(int i = 0; i < 2; i++){
        for(int j = 0; j < 2; j++){
            bool animals[N];
            animals[0] = i;
            animals[1] = j;
            for(int k = 2; k < N; k++){
                animals[k] = (responce[k - 1] ^ animals[k - 1] ^ animals[k - 2]);
            }
            if(!(animals[N - 2] ^ animals[N - 1] ^ animals[0] ^ responce[N - 1])){
                if(!(animals[N - 1] ^ animals[0] ^ animals[1] ^ responce[0])){
                    for(int k = 0; k < N; k++){
                        if(animals[k]){
                        cout << "S";
                        }else{
                            cout << "W";
                        }
                    }
                    return 0;
                }
            }
        }
    }
    cout << -1 << endl;
}
### Prompt

In Cpp, your task is to solve the following problem:

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 <iostream>
#include <vector>
using namespace std;

int main(){
    int N;
    string S;
    cin >> N >> S;
    vector<bool> responce;
    for(int i = 0; i < N; i++){
        responce.push_back(S[i] == 'o');
    }
    for(int i = 0; i < 2; i++){
        for(int j = 0; j < 2; j++){
            bool animals[N];
            animals[0] = i;
            animals[1] = j;
            for(int k = 2; k < N; k++){
                animals[k] = (responce[k - 1] ^ animals[k - 1] ^ animals[k - 2]);
            }
            if(!(animals[N - 2] ^ animals[N - 1] ^ animals[0] ^ responce[N - 1])){
                if(!(animals[N - 1] ^ animals[0] ^ animals[1] ^ responce[0])){
                    for(int k = 0; k < N; k++){
                        if(animals[k]){
                        cout << "S";
                        }else{
                            cout << "W";
                        }
                    }
                    return 0;
                }
            }
        }
    }
    cout << -1 << endl;
}
```