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:
#include <bits/stdc++.h>
#include <algorithm>
#include <unordered_map>
using namespace std;

typedef long long ll;
typedef vector<long long int> vll;
typedef vector<int> vi;
typedef unordered_map<int, int> umap;
typedef vector<bool> vb;
typedef pair<int, int> pii;
typedef vector<pii> vpii;
typedef vector<pair<int, vi>> vpiv;
typedef unordered_map<ll, vll> ull;
typedef set<ll> sll;

#define min(a, b) ((a > b) ? b : a)
#define max(a, b) ((a > b) ? a : b)
#define run(i, n) for (int i = 0; i < n; i++)
#define rund(i, n) for (int i = n - 1; i >= 0; i--)
#define in(v) insert(v)
#define pb(v) push_back(v)
#define mp(v, u) make_pair(v, u)
#define MOD 1000000007
#define lim 10000000
#define INF 1000000009
#define f first
#define s second
#define yes cout << "YES" << endl
#define no cout << "NO" << endl
//-------------------Number theory------------
// generate prime numbers till n
vb seive(int n)
{
    vb prime(n + 1, true);
    prime[0] = false;
    prime[1] = false;
    for (ll i = 2; i * i <= n; i++)
    {
        if (prime[i])
            for (ll j = i * i; j <= n; j += i)
            {
                prime[j] = false;
            }
    }
    return prime;
}

//euler totient function
vll totient(ll n)
{
    vll tot(n + 1);
    run(i, n + 1)
    {
        tot[i] = i;
    }
    for (int i = 2; i * i <= n; i++)
    {
        if (tot[i] == i)
        {
            for (int j = i; j <= n; j += i)
            {
                tot[j] /= i;
                tot[j] *= i - 1;
            }
        }
    }
    return tot;
}
//modular exponentiation
ll modExp(ll a, ll n, ll p)
{
    if (n == 1)
    {
        return a % p;
    }
    if (n % 2)
    {
        ll t = modExp(a, n / 2, p);
        return (a * t % p * t) % p;
    }
    else
    {
        ll t = modExp(a, n / 2, p);
        return t % p * t % p;
    }
}
//a inverse mod p, where a and p are coprime
ll modInv(ll a, ll p, vb prime)
{
    // we know if a and p are coprime then a^totient(p) = 1 mod p => a^{totient(p)-1}=a^-1 mod p
    vll tmp = totient(p);
    ll t = tmp[t];
    return modExp(a, t, p);
}

//----------------Segment Tree -SUM -----------------------------------------------
void buildSeg(int *a, int *t, int v, int l, int r)
{
    //initially v should be passed as 1
    //declare t in main method
    if (l == r)
    {
        t[v] = a[l];
        return;
    }
    int m = (l + r) / 2;
    buildSeg(a, t, 2 * v, l, m);         //build Left sub-tree
    buildSeg(a, t, 2 * v + 1, m + 1, r); //build Right sub-tree
    // store the sum of left and right sub-segments in the present node
    t[v] = t[2 * v] + t[2 * v + 1];
}

ll sumQuery(int *t, int v, int tl, int tr, int l, int r)
{
    // start = 0 and end = n-1 initially .
    // v initially should be 1 .
    if (l > r)
    {
        return 0;
    }
    if (tl == l && tr == r)
    {
        return t[v];
    }
    else
    {
        int tm = (tl + tr) / 2;
        if (tl <= l && l <= tm && tl <= r && r <= tm)
        {
            // the query range is completely in left sub-segment
            return sumQuery(t, 2 * v, tl, tm, l, r);
        }
        else if (tm < l && l <= tr && tm < r && r <= tr)
        {
            return sumQuery(t, 2 * v + 1, tm + 1, tr, l, r);
        }
        else
        {
            ll s1 = sumQuery(t, 2 * v, tl, tm, l, tm);
            ll s2 = sumQuery(t, 2 * v + 1, tm + 1, tr, tm + 1, r);
            return s1 + s2;
        }
    }
}

// few helpers
void print(vll a)
{
    for (int i = 0; i < a.size(); i++)
    {
        cout << a[i] << " ";
    }
    cout << endl;
}

void inp(vll &a)
{
    int i = 0;
    for (auto &it : a)
    {
        int t;
        cin >> t;
        it = t;
    }
}
class myComparator
{
public:
    bool operator()(pair<ll, ll> &p1, pair<ll, ll> &p2)
    {
        if (p1.first < p2.first)
            return true;
        else if (p1.first == p2.first)
            return p1.second > p2.second;
        else
            return false;
    }
};
class four
{
public:
    ll x1, y1, x2, y2;
};
int p1 = 1000000007;
int p2 = 1824236251;
ll p = 998244353;

//----------------MAIN METHOD (Code)------------------------------------------------
ll merge(vll& arr, vll& left, vll& right) {
    ll i = 0, j = 0, count = 0;
    while (i < left.size() || j < right.size()) {
        if (i == left.size()) {
            arr[i+j] = right[j];
            j++;
        } else if (j == right.size()) {
            arr[i+j] = left[i];
            i++;
        } else if (left[i] <= right[j]) {
            arr[i+j] = left[i];
            i++;                
        } else {
            arr[i+j] = right[j];
            count += left.size()-i;
            j++;
        }
    }
    return count;
}

ll getInvCount(vll& arr) {
    if (arr.size() < 2)
        return 0;

    int m = (arr.size() + 1) / 2;
    vll left(arr.begin() , arr.begin()+m);
    vll right(arr.begin()+ m, arr.end());

    return getInvCount(left) + getInvCount(right) + merge(arr, left, right);
}
ll findInv(string &tmp , string&s , unordered_map<char,vector<ll>>& t){
    int inv = 0;
    int c1 =0,c2 =0 , c3 = 0 , c4 =0;
    vector<ll> id;
    for(auto i:tmp ){
        if(i=='A'){
            id.pb(t['A'][c1++]);
        }
        if(i=='N'){
            id.pb(t['N'][c2++]);
        }
        if(i=='O'){
            id.pb(t['O'][c3++]);
        }
        if(i=='T'){
            id.pb(t['T'][c4++]);
        }
    }
    // print(id);
    return getInvCount(id);
}

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t ; cin>>t;
    while(t--){
        string s;cin>>s;
        unordered_map<char,int> freq;
        unordered_map<char,vll> rec;
        string ans ="";
        ll inversion_number = 0;
        vector<char> letter = {'A','N','O','T'};
        for(int i=0;i<s.length();i++){
            freq[s[i]] ++;
            rec[s[i]].pb(i);
        }
        int i = 1;
        do{
            string tmp ="";
            for(auto c : letter){
                run(i,freq[c]){
                    tmp+=c;
                }
            }
            // cout<<tmp<<endl;
            ll temp = findInv(tmp,s,rec);
            if(temp >= inversion_number){
                inversion_number = temp;
                ans = tmp;
            }
            
        }while( next_permutation(letter.begin(),letter.end()));

        cout<<ans<<endl;
    }
    
}