Problem: After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.

Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters). 

Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.

To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?

Input

The first line of input contains a single integer t (1 ≤ t ≤ 100000) — the number of testcases.

The first and only line of each testcase contains 1 string a (1 ≤ |a| ≤ 100000). a consists of only the characters "A", "N", "O" and "T".

It is guaranteed that the sum of |a| over all testcases does not exceed 100000.

Output

For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.

Example

Input


4
ANTON
NAAN
AAAAAA
OAANTTON


Output


NNOTA
AANN
AAAAAA
TNNTAOOA

Note

For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON: 

NNOTA → NNOAT → NNAOT → NANOT → NANTO → ANNTO → ANTNO → ANTON. 

Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.

For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted.

Solution (in CPP):

Solution:
/*
    Author: Nguyen Tan Bao
    Status:
    Idea:
*/

#include <bits/stdc++.h>
#define FI first
#define SE second
#define ALL(a) a.begin(), a.end()
#define SZ(a) int((a).size())
#define MS(s, n) memset(s, n, sizeof(s))
#define FOR(i,a,b) for (int i = (a); i <= (b); i++)
#define FORE(i,a,b) for (int i = (a); i >= (b); i--)
#define FORALL(it, a) for (__typeof((a).begin()) it = (a).begin(); it != (a).end(); it++)
#define TRAV(x, a) for (auto &x : a)

using namespace std;
using ll = long long; using ld = double; 
using pi = pair<int, int>; using pl = pair<ll, ll>; using pd = pair<ld, ld>;
using cd = complex<ld>; using vcd = vector<cd>;

using vi = vector<int>; using vl = vector<ll>;
using vd = vector<ld>; using vs = vector<string>;
using vpi = vector<pi>; using vpl = vector<pl>; using vpd = vector<pd>; // vector<pair>

template<class T> using min_pq = priority_queue<T, vector<T>, greater<T> >;
template<class T> inline int ckmin(T& a, const T& val) { return val < a ? a = val, 1 : 0; }
template<class T> inline int ckmax(T& a, const T& val) { return a < val ? a = val, 1 : 0; }
template<class T> void remDup(vector<T>& v) { sort(ALL(v)); v.erase(unique(ALL(v)), end(v)); }

constexpr int pct(int x) { return __builtin_popcount(x); } // # of bits set
constexpr int bits(int x) { return x == 0 ? 0 : 31-__builtin_clz(x); } // floor(log2(x)) 
constexpr int p2(int x) { return 1<<x; }
constexpr int msk2(int x) { return p2(x)-1; }

ll ceilDiv(ll a, ll b) { return a / b + ((a ^ b) > 0 && a % b); } // divide a by b rounded up
ll floorDiv(ll a, ll b) { return a / b - ((a ^ b) < 0 && a % b); } // divide a by b rounded down
void setPrec(int x) { cout << fixed << setprecision(x); }

// TO_STRING
#define ts to_string
string ts(char c) { return string(1, c); }
string ts(const char* s) { return (string) s; }
string ts(string s) { return s; }
string ts(bool b) { return ts((int)b); }
template<class T> string ts(complex<T> c) { stringstream ss; ss << c; return ss.str(); }
template<class T> using V = vector<T>;
string ts(V<bool> v) {string res = "{"; FOR(i,0,SZ(v)-1) res += char('0'+v[i]); res += "}"; return res; }
template<size_t sz> string ts(bitset<sz> b) { string res = ""; FOR(i,0,SZ(b)-1) res += char('0'+b[i]); return res; }
template<class T, class U> string ts(pair<T,U> p);
template<class T> string ts(T v) { // containers with begin(), end()
    bool fst = 1; string res = "";
    for (const auto& x: v) { if (!fst) res += " "; fst = 0; res += ts(x); }
    return res;
}
template<class T, class U> string ts(pair<T,U> p) { return "("+ts(p.FI)+", "+ts(p.SE)+")"; }

// OUTPUT
template<class T> void pr(T x) { cout << ts(x); }
template<class T, class ...U> void pr(const T& t, const U&... u) { pr(t); pr(u...); }
void ps() { pr("\n"); } // print w/ spaces
template<class T, class ...U> void ps(const T& t, const U&... u) { pr(t); if (sizeof...(u)) pr(" "); ps(u...); }

// DEBUG
void DBG() { cerr << "]" << endl; }
template<class T, class ...U> void DBG(const T& t, const U&... u) { cerr << ts(t); if (sizeof...(u)) cerr << ", "; DBG(u...); }
#define dbg(...) cerr << "Line(" << __LINE__ << ") -> [" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__)
#define chk(...) if (!(__VA_ARGS__)) cerr << "Line(" << __LINE__ << ") -> function(" \
        << __FUNCTION__  << ") -> CHK FAILED: (" << #__VA_ARGS__ << ")" << "\n", exit(0);

const ld PI = acos(-1.0);
const int dx[4] = {1,0,-1,0}, dy[4] = {0,1,0,-1};
const ld EPS = 1e-9;
const ll MODBASE = 1000000007LL;
const int INF = 0x3f3f3f3f;

const int MAXN = 100010;
const int MAXM = 1000;
const int MAXK = 16;
const int MAXQ = 200010;

struct BinaryIndexedTree {
    public:
    // 1-indexed
    BinaryIndexedTree(int n) {
        N = n;
        t.resize(N+1);
        FOR(i,0,N) t[i] = 0;
    }

    void update(int u, int val) {
        while (u <= N) {
            t[u] += val;
            u = u + (u & (-u));
        }
    }
 
    ll get(int u) {
        ll res = 0;
        while (u) {
            res += t[u];
            u = u - (u & (-u));
        }
        return res;
    }

    private:
    vector<ll> t;
    int N;
};

string s;
int n, a[MAXN];
vi f[4];

int main() {
    ios::sync_with_stdio(0);
    cin.tie(nullptr);
    int te;
    cin >> te;
    while (te--) {
        cin >> s;
        n = SZ(s);
        s = " " + s;
        FOR(i,0,3) f[i].clear();
        FOR(i,1,n) {
            if (s[i] == 'A') a[i] = 0;
            else if (s[i] == 'N') a[i] = 1;
            else if (s[i] == 'T') a[i] = 2;
            else a[i] = 3;
            f[a[i]].push_back(i);
        }
        // FOR(i,0,3) {
        //     dbg(f[i]);
        // }
        vi z = {0,1,2,3};
        vi vRes;
        ll res = -1;
        do {
            BinaryIndexedTree bit(n);
            ll sum = 0;
            FORE(i,3,0) {
                TRAV(x, f[z[i]]) {
                    sum += bit.get(x-1);
                }
                TRAV(x, f[z[i]]) {
                    bit.update(x, 1);
                }
            }
            // dbg(z, sum);
            if (sum > res) {
                res = sum;
                vRes = z;
            }
        } while (next_permutation(ALL(z)));

        FOR(i,0,3) {
            TRAV(x, f[vRes[i]]) {
                cout << s[x];
            }
        }
        cout << "\n";
    }
    return 0;
}