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.text.*;
/**
 *
 * @author alanl
 */
public class Main{
    static BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
    static StringTokenizer st;
    static PrintWriter pr = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    public static void main(String[] args) throws IOException{
        int t = readInt();
        while(t-->0){
            String s = readLine();
            int a = 0, b = 0, c = 0;
            for(int i = 0; i<s.length(); i++){
                if(s.charAt(i)=='R')a++;
                else if(s.charAt(i)=='S')b++;
                else c++;
            }
            char ans = 'a';
            if(a>=b&&a>=c)ans = 'P';
            else if(b>=a&&b>=c)ans = 'R';
            else ans = 'S';
            for(int i = 0; i<s.length(); i++)System.out.print(ans);
            System.out.println();
        }
    }
    static String next () throws IOException {
        while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(input.readLine().trim());
        return st.nextToken();
    }
    static long readLong () throws IOException {
        return Long.parseLong(next());
    }
    static int readInt () throws IOException {
        return Integer.parseInt(next());
    }
    static double readDouble () throws IOException {
        return Double.parseDouble(next());
    }
    static char readChar () throws IOException {
        return next().charAt(0);
    }
    static String readLine () throws IOException {
        return input.readLine().trim();
    }
    static void print(Object b) {
        System.out.print(b);
    }
    static void println(Object b) {
        System.out.println(b);
    }
    static void println() {
        System.out.println();
    }
    /* stuff you should look for
	* int overflow, array bounds
	* special cases (n=1?)
	* do smth instead of nothing and stay organized
	* WRITE STUFF DOWN
	* DON'T GET STUCK ON ONE APPROACH
    // Did you read the bounds?
    // Did you make typos?
    // Are there edge cases (N=1?)
    // Are array sizes proper (scaled by proper constant, for example 2* for koosaga tree)
    // Integer overflow?
    // DS reset properly between test cases?
    // Is using long longs causing TLE?
    // Are you using floating points?
    */
}
