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 bisect
import decimal
from decimal import Decimal
import os
from collections import Counter
import bisect
from collections import defaultdict
import math
import random
import heapq 
from math import sqrt
import sys
from functools import reduce, cmp_to_key
from collections import deque
import threading
from itertools import combinations
from io import BytesIO, IOBase
from itertools import accumulate

# sys.setrecursionlimit(200000)
# mod = 10**9+7
# mod = 998244353
decimal.getcontext().prec = 46

def primeFactors(n): 
    prime = set()
    while n % 2 == 0: 
        prime.add(2)
        n = n//2
    for i in range(3,int(math.sqrt(n))+1,2): 
        while n % i== 0: 
            prime.add(i)
            n = n//i 
    if n > 2: 
        prime.add(n)
    return list(prime)
    
def getFactors(n) : 
    factors = []
    i = 1
    while i <= math.sqrt(n): 
        if (n % i == 0) : 
            if (n // i == i) : 
                factors.append(i)
            else : 
                factors.append(i)
                factors.append(n//i)
        i = i + 1
    return factors
    
def modefiedSieve(): 
    mx=10**7+1
    sieve=[-1]*mx
    for i in range(2,mx):
        if sieve[i]==-1:
            sieve[i]=i
            for j in range(i*i,mx,i):
                if sieve[j]==-1:
                    sieve[j]=i
    return sieve
    
def SieveOfEratosthenes(n): 
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    num = []
    for p in range(2, n+1): 
        if prime[p]: 
            num.append(p)
    return num

def lcm(a,b):
    return (a*b)//math.gcd(a,b)

def sort_dict(key_value):
    return sorted(key_value.items(), key = lambda kv:(kv[1], kv[0]), reverse=True)

def list_input():
    return list(map(int,input().split()))
    
def num_input():
    return map(int,input().split())
    
def string_list():
    return list(input())
    
def decimalToBinary(n):  
    return bin(n).replace("0b", "")  
    
def binaryToDecimal(n): 
    return int(n,2) 
    
def DFS(n,s,adj):              
    visited = [False for i in range(n+1)]  
    stack = [] 
    stack.append(s)  
  
    while (len(stack)):  
        s = stack[-1]  
        stack.pop() 
  
        if (not visited[s]):  
            visited[s] = True 
            
        for node in adj[s]:  
            if (not visited[node]):  
                stack.append(node)  
        
def maxSubArraySum(a,size): 
    maxint = 10**10
    max_so_far = -maxint - 1
    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 lis(arr): 
    n = len(arr) 
    lis = [1]*n 
    for i in range (1 , n): 
        for j in range(0 , i): 
            if arr[i] >= arr[j] and lis[i]< lis[j] + 1 : 
                lis[i] = lis[j]+1
    maximum = 0
    for i in range(n): 
        maximum = max(maximum , lis[i]) 
  
    return maximum 
        
def solve():
    s = input()
    n = len(s)
    occur = {
        'P' : 0,
        'R' : 0,
        'S' : 0
    }
    for i in s:
        occur[i] += 1
    
    occur = sort_dict(occur)
    u,v = occur[0]
    if u == 'R':
        print('P'*n)
    elif u == 'S':
        print('R'*n)
    else:
        print('S'*n)
        
t = 1
t = int(input())
for _ in range(t):
    solve()


    
    
    
    
