#include <bits/stdc++.h>
#define For(i, a, b) for(int (i)=(a); (i)<(b); (i)++)
#define rFor(i, a, b) for(int (i)=(a)-1; (i)>=(b); (i)--)
#define rep(i, n) For((i), 0, (n))
#define rrep(i, n) rFor((i), (n), 0)
using namespace std;

char next(char now, char prev, char stat){
  if(now == 'S'){
    if((prev=='S' && stat=='o') || (prev=='W' && stat=='x')) return 'S';
    return 'W';
  }
  if((prev=='S' && stat=='x') || (prev=='W' && stat=='o')) return 'S';
  return 'W';
}

int main(){
  int n; string s; cin >> n >> s;
  string init[4] = {"SS", "SW", "WS", "WW"};
  rep(i, 4){
    string ans = init[i];
    For(j, 1, n-1) ans += next(ans[j], ans[j-1], s[j]);
    if(ans[0]==next(ans[n-1], ans[n-2], s[n-1]) && ans[1]==next(ans[0], ans[n-1], s[0])){
      cout << ans << endl;
      return 0;
    }
  }
  cout << -1 << endl;
}
### Prompt

Create a solution in Cpp for 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 <bits/stdc++.h>
#define For(i, a, b) for(int (i)=(a); (i)<(b); (i)++)
#define rFor(i, a, b) for(int (i)=(a)-1; (i)>=(b); (i)--)
#define rep(i, n) For((i), 0, (n))
#define rrep(i, n) rFor((i), (n), 0)
using namespace std;

char next(char now, char prev, char stat){
  if(now == 'S'){
    if((prev=='S' && stat=='o') || (prev=='W' && stat=='x')) return 'S';
    return 'W';
  }
  if((prev=='S' && stat=='x') || (prev=='W' && stat=='o')) return 'S';
  return 'W';
}

int main(){
  int n; string s; cin >> n >> s;
  string init[4] = {"SS", "SW", "WS", "WW"};
  rep(i, 4){
    string ans = init[i];
    For(j, 1, n-1) ans += next(ans[j], ans[j-1], s[j]);
    if(ans[0]==next(ans[n-1], ans[n-2], s[n-1]) && ans[1]==next(ans[0], ans[n-1], s[0])){
      cout << ans << endl;
      return 0;
    }
  }
  cout << -1 << endl;
}
```