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 java.io.BufferedReader;
import java.io.BufferedOutputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.List;

public class CE91B {
	public static void main(String[] args) throws NumberFormatException, IOException {
		FastReader sc = new FastReader();
		PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
	
		int t = sc.nextInt();
		
		for(int tt=0; tt<t; tt++) {
			String s = sc.next();
			char[] c = s.toCharArray();
			int rock = 0;
			int paper = 0;
			int scissors = 0;
			for(char ch: c) {
				if(ch == 'R') {
					paper++;
				} else if(ch == 'P') {
					scissors++;
				} else {
					rock++;
				}
			}
			
			int tot = rock+paper+scissors;
			
			if (rock >= paper && rock >= scissors) {
				for(int i=0; i<tot; i++) {
					out.print("R");
				}
			} else if (paper >= scissors) {
				for(int i=0; i<tot; i++) {
					out.print("P");
				}
			} else {
				for(int i=0; i<tot; i++) {
					out.print("S");
				}
			}
			
			out.println();
			
		}
		
		out.flush();
	}
	
	static class FastReader{BufferedReader br;StringTokenizer st;public FastReader()
	{br=new BufferedReader(new InputStreamReader(System.in), 32768);}
	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;}}
}
