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 <bits/stdc++.h>
using namespace std;

#define times(n, i)      uptil(0, n, i)
#define rtimes(n, i)     downto((n) - 1, 0, i)
#define upto(f, t, i)    for(int _##i = (t), i = (f); i <= _##i; i++)
#define uptil(f, t, i)   for(int _##i = (t), i = (f); i <  _##i; i++)
#define downto(f, t, i)  for(int _##i = (t), i = (f); i >= _##i; i--)
#define downtil(f, t, i) for(int _##i = (t), i = (f); i >  _##i; i--)

#define ln << endl

const int MOD = 1e9 + 7;

int main() {
    int N; cin >> N;
    string O; cin >> O;

    times(2, a) times(2, b) {
        vector<bool> S(N + 2);
        S[0] = a; S[1] = b;

        times(N, i) {
            S[i+2] = S[i] == S[i+1] == (O[i] == 'o');
        }
        if(S[0] == S[N] && S[1] == S[N+1]) {
            times(N, i) cout << (S[i+1] ? 'S' : 'W');
            cout ln;
            return 0;
        }
    }

    cout << -1 ln;

    return 0;
}
