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.*;
import java.util.*;
 
public class B{
    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()); 
        } 
        int[] readArray(int n){
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
        }
        long[] readLongArray(int n){
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    } 
    static int getgcd(int n1,int n2){
        while(n1 != n2)
        {
            if(n1 > n2)
                n1 -= n2;
            else
                n2 -= n1;
        }
        return n1;
    }
    static boolean isSorted(int[] arr, int n){
        for(int i=1;i<n;i++){
            if(arr[i]<arr[i-1]) return false;
        }
        return true;
    }
    static boolean isPrime(long n){
        if(n==2) return true;
        else if(n%2==0) return false;
        for(int i=3;i<=n/2;i+=2){
            if(n%i==0) return false;
        }
        return true;
    }
    static void printAll(char c,int n){
        for(int i=0;i<n;i++){
            System.out.print(c);
        }
        System.out.println();
    }
    public static void main(String[] args) throws NumberFormatException, IOException {
        FastReader s=new FastReader(); 
        int t=s.nextInt();
        int SC,RO,PA;
        // boolean test=true;
        String str;
        while(t>0){
            RO=0;SC=0;PA=0;
            str=s.nextLine();
            for(int i=0;i<str.length();i++){
                if(str.charAt(i)=='S') SC++;
                else if(str.charAt(i)=='R') RO++;
                else if(str.charAt(i)=='P') PA++;
            }
            int[] arr = {RO,PA,SC};
            Arrays.sort(arr);
            if(arr[2]==PA) printAll('S', str.length());
            else if(arr[2]==SC) printAll('R', str.length());
            else if(arr[2]==RO) printAll('P', str.length());
            
            t--;
        }
    }
}