Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

6
ooxoox

Sample Output 1

SSSWWS

For example, if the animals numbered 1, 2, 3, 4, 5 and 6 are respectively a sheep, sheep, sheep, wolf, wolf, and sheep, it is consistent with their responses. Besides, there is another valid assignment of species: a wolf, sheep, wolf, sheep, wolf and wolf.

Let us remind you: if the neiboring animals are of the same species, a sheep answers o and a wolf answers x. If the neiboring animals are of different species, a sheep answers x and a wolf answers o.

Sample Input 2

3
oox

Sample Output 2

-1

Print -1 if there is no valid assignment of species.

Sample Input 3

10
oxooxoxoox

Sample Output 3

SSWWSSSWWS

Solution in PyPy2 (5.6.0) (judge verdict: Runtime Error)

#include <cstdio>

constexpr int MAX_N = 100100;

using namespace std;

int n;
char arr[MAX_N];

char opt[4][MAX_N];

int main() {
    scanf(" %d", &n);
    for (int i = 0; i < n; i++) {
        scanf(" %c", &arr[i]);
    }

    opt[0][0] = 'S';
    opt[0][1] = 'S';
    opt[1][0] = 'S';
    opt[1][1] = 'W';
    opt[2][0] = 'W';
    opt[2][1] = 'S';
    opt[3][0] = 'W';
    opt[3][1] = 'W';

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < 4; j++) {
            if (opt[j][i] == 'S' && opt[j][i + 1] == 'S') {
                if (arr[i] == 'o') {
                    opt[j][i + 2] = 'S';
                } else {
                    opt[j][i + 2] = 'W';
                }
            } else if (opt[j][i] == 'S' && opt[j][i + 1] == 'W') {
                if (arr[i] == 'o') {
                    opt[j][i + 2] = 'W';
                } else {
                    opt[j][i + 2] = 'S';
                }
            } else if (opt[j][i] == 'W' && opt[j][i + 1] == 'S') {
                if (arr[i] == 'o') {
                    opt[j][i + 2] = 'W';
                } else {
                    opt[j][i + 2] = 'S';
                }
            } else {
                if (arr[i] == 'o') {
                    opt[j][i + 2] = 'S';
                } else {
                    opt[j][i + 2] = 'W';
                }
            }
        }
    }

    int which = -1;

    for (int i = 0; i < 4; i++) {
        if (opt[i][0] == opt[i][n] && opt[i][1] == opt[i][n + 1]) {
            which = i;
            break;
        }
    }

    if (which == -1) {
        printf("-1\n");
    } else {
        for (int i = 1; i <= n; i++) {
            printf("%c", opt[which][i]);
        }
        printf("\n");
    }

    return 0;
}