Recently, you found a bot to play "Rock paper scissors" with. Unfortunately, the bot uses quite a simple algorithm to play: he has a string s = s_1 s_2 ... s_{n} of length n where each letter is either R, S or P.

While initializing, the bot is choosing a starting index pos (1 ≤ pos ≤ n), and then it can play any number of rounds. In the first round, he chooses "Rock", "Scissors" or "Paper" based on the value of s_{pos}: 

  * if s_{pos} is equal to R the bot chooses "Rock"; 
  * if s_{pos} is equal to S the bot chooses "Scissors"; 
  * if s_{pos} is equal to P the bot chooses "Paper"; 



In the second round, the bot's choice is based on the value of s_{pos + 1}. In the third round — on s_{pos + 2} and so on. After s_n the bot returns to s_1 and continues his game.

You plan to play n rounds and you've already figured out the string s but still don't know what is the starting index pos. But since the bot's tactic is so boring, you've decided to find n choices to each round to maximize the average number of wins.

In other words, let's suggest your choices are c_1 c_2 ... c_n and if the bot starts from index pos then you'll win in win(pos) rounds. Find c_1 c_2 ... c_n such that (win(1) + win(2) + ... + win(n))/(n) is maximum possible.

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

Next t lines contain test cases — one per line. The first and only line of each test case contains string s = s_1 s_2 ... s_{n} (1 ≤ n ≤ 2 ⋅ 10^5; s_i ∈ \{R, S, P\}) — the string of the bot.

It's guaranteed that the total length of all strings in one test doesn't exceed 2 ⋅ 10^5.

Output

For each test case, print n choices c_1 c_2 ... c_n to maximize the average number of wins. Print them in the same manner as the string s.

If there are multiple optimal answers, print any of them.

Example

Input


3
RRRR
RSP
S


Output


PPPP
RSP
R

Note

In the first test case, the bot (wherever it starts) will always choose "Rock", so we can always choose "Paper". So, in any case, we will win all n = 4 rounds, so the average is also equal to 4.

In the second test case: 

  * if bot will start from pos = 1, then (s_1, c_1) is draw, (s_2, c_2) is draw and (s_3, c_3) is draw, so win(1) = 0; 
  * if bot will start from pos = 2, then (s_2, c_1) is win, (s_3, c_2) is win and (s_1, c_3) is win, so win(2) = 3; 
  * if bot will start from pos = 3, then (s_3, c_1) is lose, (s_1, c_2) is lose and (s_2, c_3) is lose, so win(3) = 0; 

The average is equal to (0 + 3 + 0)/(3) = 1 and it can be proven that it's the maximum possible average.

A picture from Wikipedia explaining "Rock paper scissors" game: 

<image>
import com.sun.jdi.ArrayReference;
import com.sun.source.tree.WhileLoopTree;

import java.io.*;
import java.util.*;
import java.util.concurrent.atomic.AtomicReferenceFieldUpdater;

public class CP {
    static long startTime;
    static long endTime;
    static Boolean [] prime ;
    static Scanner sc = new Scanner(System.in);
    static class FastReader{
        BufferedReader br;
        StringTokenizer st;

        public FastReader()
        {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                try
                {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException  e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt()
        {
            return Integer.parseInt(next());
        }

        long nextLong()
        {
            return Long.parseLong(next());
        }

        double nextDouble()
        {
            return Double.parseDouble(next());
        }

        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }
    public static int lowerBound
            (Integer[] array, int length, int value) {
        int l = 0;
        int r = length;
        while (l < r) {
            int mid = (l + r) / 2;
            if (value < array[mid]) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
    public static long gcd(long a, long b){
        if (b == 0)
            return a;
        else return gcd(b, a % b);
    }
    public  static  long countDivs(long n ){
        int cn=0;
        long sqr = (long)Math.sqrt(n);
        for (long i = 1; i<=sqr ; ++i) {
            if(n%i==0){
                ++cn;
            }
        }
        cn*=2;
        if(sqr*sqr==n) cn--;
        return cn;

    }
    static void prime(int  x) {
        //sieve algorithm. nlog(log(n)).
        prime=new Boolean[ (x+1)];
        Arrays.fill(prime,true);
        prime[0] = prime[1] = false;
        for (int  i = 2; i * i <= x ; i++)
            if (prime[i])
                for (int  j = i * i; j <= x; j += i)
                    prime[j] = false;
    }
    static int[] sum;
    static void cumulitiveSum( int [] arr){
        sum = new int[arr.length];
        sum[0]=arr[0];
        for (int i = 1; i <arr.length; i++) {
            sum[i]=arr[i]+sum[i-1];
        }
    }
    static boolean isEven(long x ){
        return x % 2 == 0;
    }
    static boolean isPrime(long x ){
        boolean flag = true;
        int sqr = (int)Math.sqrt(x);
        if(x<2) return false;
        for (int i = 2; i <=sqr; i++) {
            if(x%i==0){
                flag=false;
                break;
            }
        }
        return flag;
    }
    static int  mex=0;
    static int mecfi(int arr[],int n  ){
        Arrays.sort(arr);
        for (int i = 0; i <n ; i++) {
            if(arr[i]==mex) {
                mex++;
            }
        }
        return mex;
    }
    public static void main(String [] args)   {
        startTime = System.currentTimeMillis();
        int T = sc.nextInt();while (T--!=0)
        {
            solve();
        }
        endTime = System.currentTimeMillis();
        long duration= endTime-startTime;
        //System.out.println(duration);
    }
    public static void solve()  {
  //////////////////////////////////////////////////////////////////////

      String s1 =sc.next();
      char [] arr = s1.toCharArray();
      int r=0,s=0,p=0;
        for (int i = 0; i <arr.length    ; i++) {
            if(arr[i]=='R'){
               p++;

            }
            else if(arr[i]=='P'){
                s++;
            }
            else {
               r++;
            }
        }
        for (int i = 0; i < arr.length; i++) {
            if(p>=s && p>=r){
                System.out.print("P");
            }
            else if(s>=r && s>=p){
                System.out.print("S");
            }
            else {
                System.out.print("R");
            }
        }
        System.out.println();
///////////////////////////////////////////////////////////////////////
    }

}
