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.util.*;
import java.lang.Math;
import java.io.* ;

public class Account {
	
	public static void main (String[] args) {
		
//		Scanner sc= new Scanner(System.in) ;
		FastReader sc= new FastReader() ;
//		StringBuilder res= new StringBuilder() ;
		int t= sc.nextInt() ;
		while(t-- > 0) {
			String str= sc.next() ;
			char[] arr= str.toCharArray() ;
			int cntR= 0 ,cntS= 0 ,cntP= 0 ,cnt= 0 ;
			char max= '#' ,ans= '%' ;
			for(int i=0 ;i<str.length() ;i++) {
				if(arr[i] == 'R') {
					cntR++ ;
					if(cntR > cnt) {
						max= 'R' ;
						cnt = cntR ;
					}
				}else if(arr[i] == 'S') {
					cntS++ ;
					if(cntS > cnt) {
						max= 'S' ;
						cnt = cntS ;
					}
				}else {
					cntP++ ;
					if(cntP > cnt) {
						max= 'P' ;
						cnt = cntP ;
					}
				}
			}if(max == 'R') {
				ans= 'P' ;
			}else if(max == 'S') {
				ans= 'R' ;
			}else {
				ans= 'S' ;
			}
			for(int i=0 ;i<str.length() ;i++) {
				System.out.print(ans);
			}
			System.out.println();
		}
		
	}
	
//	RRSPRP
//	RRSPRP 
//	RSPRPR
//	SPRPRR
	
//	RRSP
//	RRSP -> 6 => N + ER
//	SPRR -> 6 
//	RRRR -> 8 => N*R
	
//	RRSSSPP
//	RRSSSPP
//	SSSSSSS
	
	
	public 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;
        }
 
        int [] readintarray(int n) {
            int res [] = new int [n];
            for(int i = 0; i<n; i++)res[i] = nextInt();
            return res;
        }
        long [] readlongarray(int n) {
            long res [] = new long [n];
            for(int i = 0; i<n; i++)res[i] = nextLong();
            return res;
        }
    }
	
	
}



