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.io.*;

public class cf1380_Div2_B{

  public static void main(String [] args){
	  FastScanner in = new FastScanner();
	  int t = in.nextInt();
	  for ( ; t > 0; --t) {
		  char[] arr = in.next().toCharArray();
		  char[] str = new char[arr.length];
		  int[] cnt = new int[3];
		  for (int i = 0; i < arr.length; i++) {
			  if (arr[i] == 'R')
				  cnt[0]++;
			  else if (arr[i] == 'S')
				  cnt[1]++;
			  else
				  cnt[2]++;
		  }
		  if (cnt[0] == cnt[1] && cnt[0] == cnt[2]) {
			  for (int i = 0; i < arr.length; i += 3) {
				  str[i] = 'R';
				  str[i + 1] = 'S';
				  str[i + 2] = 'P';
			  }
		  }
		  else if (cnt[0] > cnt[1] && cnt[0] > cnt[2]) {
			  for (int i = 0; i < arr.length; i++) {
				  str[i] = 'P';
			  }
		  }
		  else if (cnt[1] > cnt[2]) {
			  for (int i = 0; i < arr.length; i++) {
				  str[i] = 'R';
			  }
		  }
		  else {
			  for (int i = 0; i < arr.length; i++) {
				  str[i] = 'S';
			  }
		  }
		  System.out.println(str);
	  }
    
  }

  public static void shuffle(int[] arr) {
    	Random rgen = new Random();

    	for (int i = 0; i < arr.length; i++) {
    		int rPos = rgen.nextInt(arr.length);
    		int temp = arr[i];
    		arr[i] = arr[rPos];
    		arr[rPos]=temp;
    	}
    }

    public static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner(String s) {
            try {
                br = new BufferedReader(new FileReader(s));
            } catch (FileNotFoundException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
        }

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

        String nextToken() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    // TODO Auto-generated catch block
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        String nextLine() {
            st = null;
            try {
                return br.readLine();
            } catch (IOException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
            return null;
        }

        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(nextToken());
        }

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

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