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.*;
import java.math.*;
 
public class E
{
    public static void process(int test_number)throws IOException
    {
       	char arr[] = nln().toCharArray();
       	int r = 0, s = 0, p = 0;
       	for(char c : arr)
       		if(c == 'R')
       			r++;
       		else if(c == 'S')
       			s++;
       		else
       			p++;

       	char take = ' ';

       	if(p >= r && p >= s)
       		take = 'S';
       	else if(r >= p && r >= s)
       		take = 'P';
       	else
       		take = 'R';
       	for(int i = 1; i <= arr.length; i++)
       		p(take);
       	p("\n");
    }

	static final int oo = Integer.MAX_VALUE; 
    static final long mod = (long)1e9+7l;
    
    static FastReader sc;
    static PrintWriter out;
    public static void main(String[]args)throws IOException
    {
        out = new PrintWriter(System.out);
        sc = new FastReader();
 
        long s = System.currentTimeMillis();
        int t = 1;
        t = ni();
        for(int i = 1; i <= t; i++)
            process(i);
 
        out.flush();
        System.err.println(System.currentTimeMillis()-s+"ms");
    }

    static void trace(Object... o){ System.err.println(Arrays.deepToString(o)); };
    static void pn(Object o){ out.println(o); }
    static void p(Object o){ out.print(o); }
    static int ni()throws IOException{ return Integer.parseInt(sc.next()); }
    static long nl()throws IOException{ return Long.parseLong(sc.next()); }
    static double nd()throws IOException{ return Double.parseDouble(sc.next()); }
    static String nln()throws IOException{ return sc.nextLine(); }
    static long gcd(long a, long b)throws IOException{ return (b==0)?a:gcd(b,a%b);}
    static int gcd(int a, int b)throws IOException{ return (b==0)?a:gcd(b,a%b); }
    static int bit(long n)throws IOException{ return (n==0)?0:(1+bit(n&(n-1))); }
    
    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(); 
        } 
  
        String nextLine(){ 
            String str = ""; 
            try{ str = br.readLine(); } catch (IOException e) { e.printStackTrace(); } 
            return str; 
        } 
    } 
}
