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 math 
from collections import deque
import sys


sys.setrecursionlimit(10**4) 

def Divisors(n) : 
	
	l=[]
	i = 2
	while i <= math.sqrt(n): 
		
		if (n % i == 0) : 
			
			
			if (n // i == i) : 
				l.append(i)
			else : 
				l.append(i)
				l.append(n//i)
		i = i + 1
	return l

def SieveOfEratosthenes(n): 
	
	l=[]
	
	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
	
	for p in range(2, n+1): 
		if prime[p]: 
			l.append(p)
	
	return l		

def primeFactors(n): 
	
	l=[]
	
	while n % 2 == 0: 
		l.append(2) 
		n = n / 2
		
	for i in range(3,int(math.sqrt(n))+1,2): 
		
		while n % i== 0: 
			l.append(i) 
			n = n / i 
			
	if n > 2: 
		l.append(n)
	
	return(l)	
		

def Factors(n) : 
	
	
	result = []
	
	for i in range(2,(int)(math.sqrt(n))+1) : 

		if (n % i == 0) : 
		    
			if (i == (n/i)) : 
				result.append(i)
			else : 
			    result.append(i)
			    result.append(n//i)
		
	result.append(1)
	
	return result

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

	return max_so_far 

def longestsubarray(arr, n, k): 
    current_count = 0
      
    # this will contain length of  
    # longest subarray found 
    max_count = 0
  
    for i in range(0, n, 1): 
        if (arr[i] % k != 0): 
            current_count += 1
        else: 
            current_count = 0
        max_count = max(current_count,  
                            max_count) 
      
    return max_count 
    
#print(SieveOfEratosthenes(100))
#print(Divisors(100))
#print(primeFactors(100))
#print(Factors(100))
#print(maxSubArraySum(a))
 

def main():
    
    s=input()
    n=len(s)
    s1=""
    c1=0
    c2=0
    c3=0
    for i in range(0,len(s)):
        if s[i]=='R':
            c1+=1
        if s[i]=='S':
            c2+=1
        if s[i]=='P':
            c3+=1    
            
    if c1>=c2 and c1>=c3:
        for i in range(0,n):
            print('P',end='')
        print()    
    elif c2>=c1 and c2>=c3:
        for i in range(0,n):
            print('R',end='')
        print() 
    else:
        for i in range(0,n):
            print('S',end='')
        print() 
            
    
    
          
t=int(input())
for i in range(0,t):
    main()