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 javax.print.DocFlavor;
import javax.print.attribute.HashAttributeSet;
import java.awt.event.MouseAdapter;
import java.io.*;

import java.math.BigInteger;
import java.nio.channels.FileChannel;
import java.util.*;

import java.util.ArrayList;

import java.util.Scanner;


public class adaking {

    static class pair {
        int first;
        int second;

        pair(int a, int b) {
            first = a;
            second = b;
        }
    }

    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;
        }
    }

    static final int mod = 1000000007;

    public static int gcd(int a, int b) {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }
    public static String findSubString(String str, String pat)
    {
        int len1 = str.length();
        int len2 = pat.length();

        // check if string's length is less than pattern's
        // length. If yes then no such window can exist
        if (len1 < len2)
        {
            System.out.println("No such window exists");
            return "";
        }

        int hash_pat[] = new int[256];
        int hash_str[] = new int[256];

        // store occurrence ofs characters of pattern
        for (int i = 0; i < len2; i++)
            hash_pat[pat.charAt(i)]++;

        int start = 0, start_index = -1, min_len = Integer.MAX_VALUE;

        // start traversing the string
        int count = 0; // count of characters
        for (int j = 0; j < len1 ; j++)
        {
            // count occurrence of characters of string
            hash_str[str.charAt(j)]++;

            // If string's char matches with pattern's char
            // then increment count
            if (hash_pat[str.charAt(j)] != 0 &&
                    hash_str[str.charAt(j)] <= hash_pat[str.charAt(j)] )
                count++;

            // if all the characters are matched
            if (count == len2)
            {
                // Try to minimize the window i.e., check if
                // any character is occurring more no. of times
                // than its occurrence in pattern, if yes
                // then remove it from starting and also remove
                // the useless characters.
                while ( hash_str[str.charAt(start)] > hash_pat[str.charAt(start)]
                        || hash_pat[str.charAt(start)] == 0)
                {

                    if (hash_str[str.charAt(start)] > hash_pat[str.charAt(start)])
                        hash_str[str.charAt(start)]--;
                    start++;
                }

                // update window size
                int len_window = j - start + 1;
                if (min_len > len_window)
                {
                    min_len = len_window;
                    start_index = start;
                }
            }
        }

        // If no window found
        if (start_index == -1)
        {
            System.out.println("No such window exists");
            return "";
        }

        // Return substring starting from start_index
        // and length min_len
        return str.substring(start_index, start_index + min_len);
    }
    public static void primeFactors(long n)
    {
        // Print the number of 2s that divide n
        long ans=0;
        long m=n;
        while (n%2==0)
        {
            //System.out.print(2 + " ");
            ans=2;
            n /= 2;
        }

        // n must be odd at this point.  So we can
        // skip one element (Note i = i +2)
        for (int i = 3; i <= Math.sqrt(n); i+= 2)
        {
            // While i divides n, print i and divide n
            while (n%i == 0)
            {
               // System.out.print(i + " ");
                ans=i;
                n /= i;
            }
        }

        // This condition is to handle the case whien
        // n is a prime number greater than 2
        //System.out.println(ans);
        if (n > 2)
        {
            //System.out.print(n);
            ans=n;

        }
        if(ans==m)
            System.out.println(1+" "+(ans-1));
        else
        System.out.println(ans+" "+(m-ans));

    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        sc.nextLine();

        for (int k = 0; k < t; k++) {
            String s=sc.nextLine();
            long r=0;
            long sr=0;
            long p=0;
            for(int j=0;j<s.length();j++)
            {
                if(s.charAt(j)=='R')
                {
                    p++;
                }
                else if(s.charAt(j)=='S')
                {
                    r++;
                }
                else
                    sr++;

            }
            StringBuilder sb=new StringBuilder();
            if(sr>=r&&sr>=p)
            {
                for(int j=0;j<s.length();j++)
                    sb.append("S");

            }
            else if(r>=sr&&r>=p)
            {
                for(int j=0;j<s.length();j++)
                    sb.append("R");

            }
            else
            {
                for(int j=0;j<s.length();j++)
                    sb.append("P");
            }
            System.out.println(sb);

        }
    }
}




















