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>
/* package codechef; // don't place package name! */

import java.lang.*;
import java.io.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.util.*;
/* Name of the class has to be "Main" only if the class is public. */
public class Codechef
{
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		FastReader sc = new FastReader();
		PrintWriter out = new PrintWriter(System.out);
		int t = sc.nextInt();
		while(t-->0)
		{
		    String str = sc.next();
		    int r=0,p=0,s=0;
		    for(int i=0;i<str.length();i++)
		    {
		        if(str.charAt(i)=='R') r++;
		        else if(str.charAt(i)=='P') p++;
		        else if(str.charAt(i)=='S') s++;
		    }
		    if(r>=p && r>=s) 
		    {
		        for(int i=0;i<str.length();i++) out.print("P");
		    }
		    else if(p>=r && p>=s)
		    {
		        for(int i=0;i<str.length();i++) out.print("S");
		    }
		    else if(s>=p && s>=r)
		    {
		        for(int i=0;i<str.length();i++) out.print("R");
		    }
		    else
		    {
		        for(int i=0;i<str.length();i++) out.print("P");
		    }
		    out.println();
		}
		
		
		out.close();
	}
}

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; 
    } 
}