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 os
import sys
from io import BytesIO, IOBase
#import sys
#input = sys.stdin.buffer.readline
BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#######################################
from itertools import permutations
from collections import *
from collections import deque
from operator import itemgetter , attrgetter
from decimal import *
import bisect
import math
import heapq as hq
#import sympy
MOD=10**9 +7


def is_prime(n):
  if n == 2 or n == 3: return True
  if n < 2 or n%2 == 0: return False
  if n < 9: return True
  if n%3 == 0: return False
  r = int(n**0.5)
  # since all primes > 3 are of the form 6n ± 1
  # start with f=5 (which is prime)
  # and test f, f+2 for being prime
  # then loop by 6.
  f = 5
  while f <= r:

    if n % f == 0: return False
    if n % (f+2) == 0: return False
    f += 6
  return True

def pow(a,b,m):
    ans=1
    while b:
        if b&1:
            ans=(ans*a)%m
        b//=2
        a=(a*a)%m
    return ans
#vis=[]
#graph=[]

def ispalindrome(s):
    if s[:]==s[::-1]:
        return 1
    return 0
dp=[]
limit=[]
v=[]
def dpdfs(u,t=-1):
    dp[0][u]=0
    dp[1][u]=0
    for i in v[u]:
        if i==t:

            continue
        if dp[1][i]==-1:
            dpdfs(i,u)
        dp[0][u]+=max(abs(limit[0][u]-limit[1][i])+dp[1][i],abs(limit[0][u]-limit[0][i])+dp[0][i])
        dp[1][u] += max(abs(limit[1][u] - limit[1][i]) + dp[1][i], abs(limit[1][u] - limit[0][i]) + dp[0][i])
vis=[]
f=0
def dfs(i):
    vis[i]=1
    act[i]=1
    for j in v[i]:
        if act[j]:
            f=1
            #print(-1)
            return -1
        if vis[j]==0:
           if dfs(j)==-1:
               return -1
    act[i]=0
    ans.append(i)
    return 0
from queue import PriorityQueue

def z_algorithm(s):
    res = [0] * len(s)
    res[0] = len(s)
    i, j = 1, 0
    while i < len(s):
        while i + j < len(s) and s[j] == s[i + j]:
            j += 1
        res[i] = j
        if j == 0:
            i += 1
            continue
        k = 1
        while i + k < len(s) and k + res[k] < j:
            res[i + k] = res[k]
            k += 1
        i, j = i + k, j - k
    return res
def gcd(a, b):
    if a == 0:
        return b
    return gcd(b % a, a)


# Function to return LCM of two numbers
def lcm(a, b):
    return (a / gcd(a, b)) * b


def gcdExtended(a, b):
    # Base Case
    if a == 0:
        return b, 0, 1

    gcd, x1, y1 = gcdExtended(b % a, a)

    # Update x and y using results of recursive
    # call
    x = y1 - (b // a) * x1
    y = x1

    return gcd, x, y
import re

for _ in range(int(input())):
    cntP=0
    cntR=0
    cntS=0
    s=input()
    l=[]
    for i in s:
        if i=="P":
            cntS+=1
        elif i=="R":
            cntP+=1
        else:
            cntR+=1
    if cntP>=max(cntR,cntS):
        print(len(s)*"P")
    elif cntS>=max(cntP,cntR):
        print(len(s)*"S")
    else:
        print(len(s)*"R")













