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 C (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

int descending_compare(const void *a, const void *b){
    if (*(long long*)a > *(long long*)b){
        return -1;
    }else if (*(long long*)a == *(long long*)b){
        return 0;
    }else{
        return 1;
    }
}

int ascending_compare(const void *a, const void *b){
    if (*(int*)a < *(int*)b){
        return -1;
    }else if (*(int*)a == *(int*)b){
        return 0;
    }else{
        return 1;
    }
}


int lower_bound(int *a, int n, int key){
    int left, mid, right;
    left = 0, right = n;
    mid = (left + right)/2;
    while ((left+1 != mid || mid+1 != right) && mid != left){
        if (key > a[mid]){
            left = mid;
        }else{
            right = mid+1;
        }
        mid = (left + right)/2;
    }
    if (a[left] >= key)return left;
    if (a[mid] >= key)return mid;
    if (a[right] >= key)return right;
    return n;
}

//greatest common divisor
unsigned long gcd(unsigned long x, unsigned long y){
    if (y == 0){ 
        return x;
    }else if (x > y){
        return gcd(y, x % y);
    }else{
        return gcd(x, y % x);
    }
}

unsigned long lcm(unsigned long x, unsigned long y){
    unsigned long g = gcd(x, y);
    return x*y/g;

}



long long factorial(int x){
    long long rtn = 1;
    int i;
    for (i = x; i > 1; i--){
        rtn = (rtn*i);
    }
    return rtn;
}

/*
int struct_ascending_compare(const void *p, const void *q) {
return ((struct_name*)p)->member_name - ((struct_name*)q)->member_name;
}*/

/*unsigned long long pascal[100][100] = {0};
void make_pascal(void){
    for (int i = 0; i < 100; i++){
        pascal[i][0] = 1;
    }
    pascal[1][1] = 1;
    for (int i = 2; i < 100; i++){
        for (int j = 1; j < 100; j++){
           pascal[i][j] = (pascal[i-1][j-1]+pascal[i-1][j]);
        }
    }
}*/
long long mod = 1000000007;

long long mod_pow(long long x, long long n){
    long long res = 1;
    for(int i = 0;i < 60; i++){
        if(n >> i & 1) res = res * x % mod;
        x = x * x % mod;
    }
    return res;
}

char s[100005];
int animal[100005] = {0};
int main(void){
    int n;
    scanf("%d", &n);
    scanf(" %s", s);
    for (int i = 0; i < n; i++){
        if (i == 0){
            if (s[i] == 'o') animal[i+1] = animal[i-1];
            else animal[i+1] = animal[i-1] ^ 1;
            continue;         
        }


        if (animal[i]){
            if (s[i] == 'x') animal[i+1] = animal[i-1];
            else animal[i+1] = animal[i-1] ^ 1;
        }else{
            if (s[i] == 'o') animal[i+1] = animal[i-1];
            else animal[i+1] = animal[i-1] ^ 1;
        }
    }

    if (animal[0] != animal[n]){
        printf("-1\n");
        return 0;
    }else{
        for (int i = 0; i < n; i++){
            if (animal[i]){
                printf("W");
            }else{
                printf("S");
            }
        }
        printf("\n");
    }
    return 0;
}