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>
from bisect import *
from collections import *
from itertools import *
import functools
import sys
import math
from decimal import *
from copy import *
from heapq import *
from fractions import *
getcontext().prec = 30
MAX = sys.maxsize
MAXN = 1000010
MOD = 10**9+7
spf = [i for i in range(MAXN)]
spf[0]=spf[1] = -1
def sieve():
    for i in range(2,MAXN,2):
        spf[i] = 2
    for i in range(3,int(MAXN**0.5)+1):
        if spf[i]==i:
            for j in range(i*i,MAXN,i):
                if spf[j]==j:
                    spf[j]=i
def fib(n,m):
    if n == 0:
        return [0, 1]
    else:
        a, b = fib(n // 2)
        c = ((a%m) * ((b%m) * 2 - (a%m)))%m
        d = ((a%m) * (a%m))%m + ((b)%m * (b)%m)%m
        if n % 2 == 0:
            return [c, d]
        else:
            return [d, c + d]
 
def charIN(x= ' '):
    return(sys.stdin.readline().strip().split(x))
 
def arrIN(x = ' '):
    return list(map(int,sys.stdin.readline().strip().split(x)))
 
def ncr(n,r):
    num=den=1
    for i in range(r):
        num = (num*(n-i))%MOD
        den = (den*(i+1))%MOD
 
    return (num*(pow(den,MOD-2,MOD)))%MOD
 
def flush():
    return sys.stdout.flush()
 
'''*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*'''
for _ in range(int(input())):
    s = list(input())
    x = [s.count('R'),s.count('P'),s.count('S')]
    idx = x.index(max(x))
    if idx==0:
        print('P'*len(s))
    elif idx==1:
        print('S'*len(s))
    else:
        print('R'*len(s))