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>
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void show(const T& v) {
  cerr << v << ' ';
}
template <typename T1, typename... T2>
void show(const T1& first, const T2&... rest) {
  show(first);
  show(rest...);
}
template <typename F, typename S>
ostream& operator<<(ostream& os, const pair<F, S>& p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "{";
  typename vector<T>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v) {
  os << "[";
  typename set<T>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename F, typename S>
ostream& operator<<(ostream& os, const map<F, S>& v) {
  os << "[";
  typename map<F, S>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
const double eps = 1e-9;
void solve() {
  string str;
  cin >> str;
  int r, s, p;
  r = s = p = 0;
  for (auto& val : str) {
    if (val == 'R')
      r++;
    else if (val == 'S')
      s++;
    else
      p++;
  }
  vector<pair<int, char> > freq;
  freq.push_back(make_pair(r, 'P'));
  freq.push_back(make_pair(p, 'S'));
  freq.push_back(make_pair(s, 'R'));
  sort(freq.rbegin(), freq.rend());
  string ans = "";
  for (int i = 0; i < str.size(); i++) {
    ans += freq[0].second;
  }
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  int t = 1, cs = 0;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
