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;
using db = double;
using ll = long long;
using ull = unsigned long long;
using pII = pair<int, int>;
using pLL = pair<ll, ll>;
constexpr int mod = 1e9 + 7;
template <class T1, class T2>
inline void chadd(T1 &x, T2 y, int Mod = mod) {
  x += y;
  while (x >= Mod) x -= Mod;
  while (x < 0) x += Mod;
}
template <class T1, class T2>
inline void chmax(T1 &x, T2 y) {
  if (x < y) x = y;
}
template <class T1, class T2>
inline void chmin(T1 &x, T2 y) {
  if (x > y) x = y;
}
inline int nextInt() {
  int x;
  cin >> x;
  return x;
}
void rd() {}
template <class T, class... Ts>
void rd(T &arg, Ts &...args) {
  cin >> arg;
  rd(args...);
}
void err() {
  cout << "\033[39;0m"
       << "\n";
}
template <class T, class... Ts>
void err(const T &arg, const Ts &...args) {
  cout << arg << ' ';
  err(args...);
}
template <template <typename...> class T, typename t, typename... A>
void err(const T<t> &arg, const A &...args) {
  for (auto &v : arg) cout << v << ' ';
  err(args...);
}
void ptt() { cout << "\n"; }
template <class T, class... Ts>
void ptt(const T &arg, const Ts &...args) {
  cout << ' ' << arg;
  ptt(args...);
}
template <class T, class... Ts>
void pt(const T &arg, const Ts &...args) {
  cout << arg;
  ptt(args...);
}
void pt() {}
template <template <typename...> class T, typename t, typename... A>
void pt(const T<t> &arg, const A &...args) {
  for (int i = 0, sze = arg.size(); i < sze; ++i)
    cout << arg[i] << " \n"[i == sze - 1];
  pt(args...);
}
inline ll qpow(ll base, ll n) {
  assert(n >= 0);
  ll res = 1;
  while (n) {
    if (n & 1) res = res * base % mod;
    base = base * base % mod;
    n >>= 1;
  }
  return res;
}
constexpr int N = 1e5 + 10;
int n;
string s;
void run() {
  cin >> s;
  map<char, int> mp;
  mp['R'] = mp['S'] = mp['P'] = 0;
  for (auto &ch : s) ++mp[ch];
  string t = "RSP", _t = "PRS";
  for (int i = 0; i < 3; ++i) {
    int ok = 1;
    for (int j = 0; j < 3; ++j)
      if (i != j) {
        if (mp[t[j]] > mp[t[i]]) ok = 0;
      }
    if (ok) {
      string res(((int)(s).size()), _t[i]);
      cout << res << "\n";
      return;
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout << fixed << setprecision(20);
  int _T = nextInt();
  while (_T--) run();
  return 0;
}
