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,io
from sys import stdout
import collections
# import random
# import math
# from operator import itemgetter
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
# from collections import Counter
# from decimal import Decimal
# import heapq
# from functools import lru_cache

import sys
# import threading
# sys.setrecursionlimit(10**6)
# threading.stack_size(102400000)

# from functools import lru_cache
# @lru_cache(maxsize=None)

######################
# --- Maths Fns  --- #
######################
def primes(n):
    sieve = [True] * n
    for i in range(3,int(n**0.5)+1,2):
        if sieve[i]:
            sieve[i*i::2*i]=[False]*((n-i*i-1)//(2*i)+1)
    return [2] + [i for i in range(3,n,2) if sieve[i]]

def binomial_coefficient(n, k):
    if 0 <= k <= n:
        ntok = 1
        ktok = 1
        for t in range(1, min(k, n - k) + 1):
            ntok *= n
            ktok *= t
            n -= 1
        return ntok // ktok
    else:
        return 0

def powerOfK(k, max):
    if k == 1:
        return [1]
    if k == -1:
        return [-1, 1]

    result = []
    n = 1
    while n <= max:
        result.append(n)
        n *= k
    return result

def divisors(n):
    i = 1
    result = []
    while i*i <= n:
        if n%i == 0:
            if n/i == i:
                result.append(i)
            else:
                result.append(i)
                result.append(n/i)
        i+=1
    return result

def distance(xa, ya, xb, yb):
    return ((ya-yb)**2 + (xa-xb)**2)**(0.5)

# @lru_cache(maxsize=None)
def digitsSum(n):
    if n == 0:
        return 0
    r = 0
    while n > 0:
        r += n % 10
        n //= 10
    return r

######################
# ---- GRID Fns ---- #
######################
def isValid(i, j, n, m):
    return i >= 0 and i < n and j >= 0 and j < m

def print_grid(grid):
    for line in grid:
        print(" ".join(map(str,line)))

######################
# ---- MISC Fns ---- #
######################
def kadane(a,size):
    max_so_far = 0
    max_ending_here = 0

    for i in range(0, size):
        max_ending_here = max_ending_here + a[i]
        if (max_so_far < max_ending_here):
            max_so_far = max_ending_here

        if max_ending_here < 0:
            max_ending_here = 0
    return max_so_far

def prefixSum(arr):
    for i in range(1, len(arr)):
        arr[i] = arr[i] + arr[i-1]
    return arr

def ceil(n, d):
    if n % d == 0:
        return n // d
    else:
        return (n // d) + 1

def _abs(a, b):
    return int(abs(a - b))

def evenodd(l):
    even = [e for e in l if e & 1 == 0]
    odd = [e for e in l if e & 1]
    return (even, odd)

# INPUTS  --------------------------
# s = input().decode('utf-8').strip()
# n = int(input())
# l = list(map(int, input().split()))
# t = int(input())
# for _ in range(t):
from collections import Counter
t = int(input())
map = {
    'R':'P',
    'P':'S',
    'S':'R',
}
for _ in range(t):
    s = input().decode('utf-8').strip()
    c = Counter(s)
    m = c.most_common()[0][0]
    print(map[m]*len(s))
