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    : raj1307 - Raj Singh
 * Date      : 28.05.2021
 */
 
#pragma GCC optimize("Ofast")
// #pragma GCC optimize("unroll-loops")
// #define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
using namespace std;

// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;
 
#define fio ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define fr(i, a, b) for(int i = (a); i < (b); ++i)
#define fb(i, b, a) for(int i = (b); i > (a); --i)
#define fill(x,y) memset(x,y,sizeof(x))
#define rr return 
#define mp make_pair
#define pb push_back
#define sz(x) ((int)x.size())
#define int long long
#define ll long long
#define ld long double
#define S second
#define F first
#define endl "\n"
#define Endl "\n"
#define all(v) (v).begin(), (v).end()
typedef pair<int, int> pii;
int max(int a,int b){if(a>b) return a; else return b;}
int min(int a,int b){if(a>b) return b; else return a;}
int gcd(int a,int b){ return (b == 0) ? a : gcd(b, a % b);}
inline int lcm(int a,int b){ return a / gcd(a, b) * b;}
inline int normalize(int x,int mod){ x %= mod; if (x < 0) x += mod; return x;}
 
int dx[4]={0,1,0,-1};
int dy[4]={1,0,-1,0};
string abc="abcdefghijklmnopqrstuvwxyz";
const double PI=acosl(-1);
const int mod = 1000*1000*1000+7;
//const int mod = 998244353; 
const int inf = 1e18+5;
 
//a and b are assumed to be taken modulo p
inline int add(int a, int b, int p = mod){ int c = a + b; if(c >= p) c -= p; return c; } // expects c doesn't exceeds 2p
inline int sub(int a, int b, int p = mod){ int c = a - b; if(c < 0) c += p; return c; } // ...  
inline int mul(int a, int b, int p = mod){ return (a * 1LL * b) % p; }  // ...

// template<typename T>
// using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>; 
//1. order_of_key(k) : number of elements strictly lesser than k
//2. find_by_order(k) : k-th element in the set
//To make it an ordered_multiset, use pairs of (value, time_of_insertion)
//to distinguish values which are similar.
 
int floor1(int n,int k){
    if(n%k == 0 || n >= 0)return n/k;
    return (n/k)-1;
}
 
int ceil1(int n,int k){
    return floor1(n+k-1,k);
}
 
int powm(int a, int b) {
    int res=1;
    while(b) {
        if(b&1)
            res=(res*a)%mod;
        a=(a*a)%mod;
        b>>=1;
    }
    return res;
}
 
 
const int N = 1e5 + 5;



int n;
string s;
int bit[N];
vector<int>g[26];
vector<int>p[26];

void update(int idx, int val)  // 1 - indexed
{
    while(idx<=n)
    {
        bit[idx]+=val;
        idx+=idx&-idx;
    }
}

int pref(int idx)
{
    int ans=0;
    while(idx>0)
    {
        ans+=bit[idx];
        idx-=idx&-idx;
    }
    return ans;
}

int rsum(int l, int r)
{
    return pref(r) - pref(l-1);
}


void solve(){

	cin>>s;
	n=sz(s);
	s=" "+s;
	fr(i,0,n+5) bit[i]=0;
	fr(i,0,26) g[i].clear();
	fr(i,1,n+1) g[s[i]-'A'].pb(i);


	vector<int>f;
	f.pb('A'-'A');
	f.pb('N'-'A');
	f.pb('O'-'A');
	f.pb('T'-'A');


	vector<int>ans=f;
	int mx=0;

	
	do{
		// f.clear();
		// f.pb('O'-'A');
		// f.pb('N'-'A');
		// f.pb('A'-'A');
		// f.pb('T'-'A');

		fr(i,0,26) p[i].clear();
		fr(i,0,n+5) bit[i]=0;
		string x=" ";
		int cnt=1;
		fr(i,0,4){
			fr(j,0,sz(g[f[i]])){
				x+=('A'+f[i]);
				p[f[i]].pb(cnt);
				++cnt;
			}
		}

		fr(i,1,n+1){
			p[x[i]-'A'].pb(i);
		}

		fr(i,0,26) reverse(all(p[i]));

		cnt=0;

		//cout<<x<<endl;
		
		fr(i,1,n+1){
			int nxt=p[s[i]-'A'].back();
			p[s[i]-'A'].pop_back(); // forgot
			//cout<<i<<" "<<nxt<<" "<<pref(nxt)<<" "<<abs(pref(nxt)+nxt-i)<<endl;


			cnt+=abs(pref(nxt)+nxt-i);
			// if(i<=n){
			// 	cnt+=abs(pref(nxt)+nxt-i);
			// 	update(i-pref(nxt),1);
			// 	update(nxt,-1);
			// }

			update(1,1);
			update(nxt+1,-1);
			
		}

		if(cnt>mx){
			ans=f;
			mx=cnt;

		}



	}while(next_permutation(all(f)));


	string x="";
	int cnt=1;
	fr(i,0,4){
		fr(j,0,sz(g[ans[i]])){
			x+=('A'+ans[i]);
			++cnt;
		}
	}

	cout<<x<<endl;
	
	



	

} 

 
signed main() {
    fio;
    #ifndef ONLINE_JUDGE 
        freopen("input.txt", "r", stdin); 
        freopen("output.txt", "w", stdout); 
        //freopen("error.txt", "w", stderr); 
    #endif 
    cout<<fixed<<setprecision(10);
    int t=1;
    cin>>t;  // UNCOMMENT FOR TEST-CASES
    //precompute();
    fr(i,1,t+1){
        solve();
    }   
    return 0;
}