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.BigInteger;

public class Solution {

    static class Pair<A, B> {

        A parent;
        B rank;

        Pair(A parent, B rank) {
            this.rank = rank;
            this.parent = parent;
        }
    }
    static class Node {
        int r;
        int c;
        int value;
        Node(int i,int j) {
            r=i;
            c=j;
            value=0;
        }
    }
    static int m=1000000007;
    static ArrayList<Integer> graph[];
    public static void main(String[] args) throws IOException {
        FastReader s1 = new FastReader();
        StringBuilder sb = new StringBuilder();
        int t=s1.I();
        while(t--!=0)
        {
            char ar[] =s1.next().toCharArray();
            int n=ar.length;
            int dp[] = new int[3];
            for(char c:ar)
            {
                if(c=='R')
                    dp[0]++;
                else if(c=='P')
                    dp[1]++;
                else
                    dp[2]++;
            }
            int max=0;
            for(int i=1;i<3;i++)
            {
                if(dp[i]>dp[max])
                    max=i;
            }
            char temp;
            if(max==0)
                temp='P';
            else if(max==1)
                temp='S';
            else
                temp='R';
            for(int i=0;i<n;i++)
                sb.append(temp);
            sb.append("\n");
        }
        System.out.println(sb);
    }
    static int lis(ArrayList<Integer> ar,int i)
    {
        int n=ar.size();
        if(i==n-1)
            return 1;
        if(ar.get(i)<ar.get(i+1))
        {
            int count=1;
            while(i<n-1 && ar.get(i)<ar.get(i+1) && count<4)
            {
                i++;
                count++;
            }
            return count;
        }
        else
        {
            int count=1;
            while(i<n-1 && ar.get(i)>ar.get(i+1) && count<4)
            {
                i++;
                count++;
            }
            return -count;
        }
    }
    static long lcm(int a,int b)
    {
        long h=gcd(a,b);
        long ans=(a/h)*(b/h);
        ans=ans*h;
        return ans;
    }
    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 I() {
            return Integer.parseInt(next());
        }

        long L() {
            return Long.parseLong(next());
        }

        double D() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    static long gcd(long a, long b) {
        if (a % b == 0) {
            return b;
        }
        return gcd(b, a % b);
    }

    static float power(float x, int y) {
        float temp;
        if (y == 0) {
            return 1;
        }
        temp = power(x, y / 2);

        if (y % 2 == 0) {
            return temp * temp;
        } else {
            if (y > 0) {
                return x * temp * temp;
            } else {
                return (temp * temp) / x;
            }
        }
    }
    static long pow(long x, long y) {
        long res = 1;
        x = x % m;
        if (x < 0) {
            x += m;
        }
        while (y > 0) {
            if ((y & 1) == 1) {
                res = (res * x) % m;
                if (res < 0) {
                    res += m;
                }
            }
            y = y >> 1;
            x = (x * x) % m;
            if (x < 0) {
                x = x + m;
            }
        }
        res = res % m;
        if (res < 0) {
            res += m;
        }
        return res;
    }

    static void sieveOfEratosthenes(int n) {
        ArrayList<Integer> prime = new ArrayList<Integer>();
        boolean Prime[] = new boolean[n + 1];
        for (int i = 2; i < n; i++) {
            Prime[i] = true;
        }

        for (int p = 2; p * p <= n; p++) {
            if (Prime[p] == true) {
                prime.add(p);
                for (int i = p * p; i <= n; i += p) {
                    Prime[i] = false;
                }
            }
        }
    }
}