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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <cmath>
#include <algorithm>
#include <vector>
#include <cstring>
#include <deque>
#include <stack>
#include <stdio.h>
#include <map>
#include <set>
#include <time.h>
#include <string>
#include <fstream>
#include <queue>
#include <bitset>
#include <cstdlib>
#include <assert.h>
#define X first
#define Y second
#define mp make_pair
#define pb push_back
#define pdd pair<double,double>
#define pii pair<ll,ll>
#define PI 3.14159265358979323846
#define MOD 1000000007
#define MOD2 1000000009
#define INF ((ll)1e+18)
#define x1 fldgjdflgjhrthrl
#define x2 fldgjdflgrtyrtyjl
#define y1 fldggfhfghjdflgjl
#define y2 ffgfldgjdflgjl
#define N 131072
#define SUM 377
#define MAG 33554431
#define RED 0
#define BLUE 1
#define eps 1e-7
typedef long long ll;
typedef long double ld;
using namespace std;
ll i,j,n,k,l,m,tot, flag,r,z, K,x1,y1,x2,y2,x3,y3,x,y,h,num,h2,timer,sz,q,ans, curL, curR, SQ, P, Q, bad;
string s, t, s1, t1;
string tran(string x)
{
    for (int i = 0; i < x.size(); i++)
        if (x[i] == 0)
            x[i] = 'S';
        else
            x[i] = 'W';
    return x;
}
int main() {
	//freopen("input.txt","r",stdin);
 	//freopen("output.txt","w",stdout);
 	cin >> n;
 	cin >> s;
 	t = s;
 	for (t[0] = 0; t[0] <= 1; t[0]++)
        for (t[1] = 0; t[1] <= 1; t[1]++)
        {
            for (i = 2; i < n; i++)
            {
                if (s[i-1] == 'x')
                    t[i] = (1^t[i-2]^t[i-1]);
                else
                    t[i] = (t[i-1]^t[i-2]);
            }
            //cout << tran(t) << endl;
            if ((s[0] == 'x' && (t[n-1]^t[0]^1) == t[1] || s[0] == 'o' && (t[n-1]^t[0]) == t[1])&&
                (s[n-1] == 'x' && (t[n-2]^t[n-1]^1) == t[0] || s[n-1] == 'o' && (t[n-2]^t[n-1]) == t[0]))
            {
                cout << tran(t) << endl;
                return 0;
            }
        }
    cout << -1 << endl;
 	return 0;
}