///by Saton.
#include<bits/stdc++.h>
#define PI acos(-1)
#define fi first
#define se second 
#define sz(a) ((int)a.size())
#define all(a) a.begin(), a.end()
#define LL long long
#define ll __int128
#define DD double double
#define LD long double
#define rep(i,a,b) for(LL i = (a);i <= (b);i ++)
#define lep(i,a,b) for(LL i = (a);i >= (b);i --)
#define FLUSH fflush(stdout)
using namespace std;
const int N = 2e5 + 10,mod =  1e9+7,P = 131;
const LL inf = 1e9+10,INF = 1e18+10;
typedef pair<int,int> PII;
typedef pair<LL,int> PLI;
typedef pair<LL,LL> PLL;
//__builtin_popcountll()计算整数二进制表示中1的个数
// mt19937_64 rnd(random_device{}());
// uniform_int_distribution<LL> dist(0, LLONG_MAX);//dist(rnd)生成随机数
LL n,m,k;
string str;

bool cmp(char x,char y) {
    if(x=='S' && y=='P') return true;
    if(x=='P' && y=='R') return true;
    if(x=='R' && y=='S') return true;
    return false;
}

void solve() {
    cin >> str;
    n = sz(str);
    str = " " + str;
    
    stack<int> stk;
    rep(i,1,n) {
        while(!stk.empty() && !cmp(str[stk.top()],str[i])) stk.pop();
        stk.push(i);
    }
    
    string ans;
    while(!stk.empty()) {
        ans = str[stk.top()];
        stk.pop();
    }
    
    cout << ans << '\n';
} 

int main() {
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int T;
    cin >> T;
    while(T --) { 
        solve();
    }
    // solve();
       
    return 0;
}  
/*   /\_/\
*   (= ._.)
*   / >  \>
*/