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 <vector>
#define rep(i, n) for (int i = 0; i < (n); i++)
using namespace std;

bool getAnswer(const string& S, string& T) {
  int N = S.size();
  for (int i = 1; i < N; ++i) {
    if ((S[i] == 'o' && T[i] == 'S') || (S[i] == 'x' && T[i] == 'W')) {
      T += T[i - 1];
    } else {
      T += (T[i - 1] == 'S') ? 'W' : 'S';
    }
  }
  if (T[0] != T.back()) return false;
  if ((S[0] == 'o' && T[0] == 'S') || (S[0] == 'x' && T[0] == 'W')) {
    if (T[1] != T[N - 1]) return false;
  } else {
    if (T[1] == T[N - 1]) return false;
  }
  return true;
}

int main(void) {
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  string S;
  cin >> S;

  const vector<string> pattern = {"SS", "SW", "WS", "WW"};
  rep(i, (int)pattern.size()) {
    string answer = pattern[i];
    if (getAnswer(S, answer)) {
      cout << answer.substr(0, N) << endl;
      return 0;
    }
  }
  cout << -1 << endl;
  return 0;
}
