Problem:
Score : 1100 points

Problem Statement

Let f(A, B), where A and B are positive integers, be the string satisfying the following conditions:

f(A, B) has length A + B;

f(A, B) contains exactly A letters A and exactly B letters B;

The length of the longest substring of f(A, B) consisting of equal letters (ex., AAAAA or BBBB) is as small as possible under the conditions above;

f(A, B) is the lexicographically smallest string satisfying the conditions above.

For example, f(2, 3) = BABAB, and f(6, 4) = AABAABAABB.

Answer Q queries: find the substring of f(A_i, B_i) from position C_i to position D_i (1-based).

Constraints

1 \leq Q \leq 10^3

1 \leq A_i, B_i \leq 5 \times 10^8

1 \leq C_i \leq D_i \leq A_i + B_i

D_i - C_i + 1 \leq 100

All input values are integers.

Partial Score

500 points will be awarded for passing the testset satisfying 1 \leq A_i, B_i \leq 10^3.

Input

Input is given from Standard Input in the following format:

Q
A_1 B_1 C_1 D_1
A_2 B_2 C_2 D_2
:
A_Q B_Q C_Q D_Q

Output

For each query i in order of input, print a line containing the substring of f(A_i, B_i) from position C_i to position D_i (1-based).

Sample Input 1

5
2 3 1 5
6 4 1 10
2 3 4 4
6 4 3 7
8 10 5 8

Sample Output 1

BABAB
AABAABAABB
A
BAABA
ABAB

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

//{
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef double lf;
typedef pair<ll,ll> ii;
#define REP(i,n) for(ll i=0;i<n;i++)
#define REP1(i,n) for(ll i=1;i<=n;i++)
#define FILL(i,n) memset(i,n,sizeof i)
#define X first
#define Y second
#define SZ(_a) (int)_a.size()
#define ALL(_a) _a.begin(),_a.end()
#define pb push_back
#ifdef brian
#define debug(...) do{\
    fprintf(stderr,"%s - %d (%s) = ",__PRETTY_FUNCTION__,__LINE__,#__VA_ARGS__);\
    _do(__VA_ARGS__);\
}while(0)
template<typename T>void _do(T &&_x){cerr<<_x<<endl;}
template<typename T,typename ...S> void _do(T &&_x,S &&..._t){cerr<<_x<<" ,";_do(_t...);}
template<typename _a,typename _b> ostream& operator << (ostream &_s,const pair<_a,_b> &_p){return _s<<"("<<_p.X<<","<<_p.Y<<")";}
template<typename It> ostream& _OUTC(ostream &_s,It _ita,It _itb)
{
    _s<<"{";
    for(It _it=_ita;_it!=_itb;_it++)
    {
        _s<<(_it==_ita?"":",")<<*_it;
    }
    _s<<"}";
    return _s;
}
template<typename _a> ostream &operator << (ostream &_s,vector<_a> &_c){return _OUTC(_s,ALL(_c));}
template<typename _a> ostream &operator << (ostream &_s,set<_a> &_c){return _OUTC(_s,ALL(_c));}
template<typename _a,typename _b> ostream &operator << (ostream &_s,map<_a,_b> &_c){return _OUTC(_s,ALL(_c));}
template<typename _t> void pary(_t _a,_t _b){_OUTC(cerr,_a,_b);cerr<<endl;}
#define IOS()
#else
#define debug(...)
#define pary(...)
#define endl '\n'
#define IOS() ios_base::sync_with_stdio(0);cin.tie(0);
#endif // brian
//}


const ll MAXn=1e5+5,MAXlg=__lg(MAXn)+2;
const ll MOD=1000000007;
const ll INF=ll(1e15);

int main()
{
    IOS();
    ll Q;
    cin>>Q;
    while(Q--)
    {
      ll n,m,l,r;
      cin>>n>>m>>l>>r;
      ll k = (max(n,m)-1) / (min(n,m) + 1) + 1;
      debug(k);
      ll mx = 0;
      // ki +  j + p = n
      //  i + kj + q = m
      for(int i = 0;i * k <= n && i <= m;i++)
      {
        ll j = (m - i) / k;
        ll p = n - k * i - j;
        ll q = m - i - k * j;
        if(p >= 0 && q >= 0)
        {
          debug(i);
          mx = i;
        }
      }
      string s = "";
      ll i = mx;
      ll j = (m - i) / k;
      ll p = n - k * i - j;
      ll q = m - i - k * j;
      if(p == -1)
      {
        p = 0;
        q += k;
        j--;
      }
      debug(mx,i,j,p,q);
      REP(x,i)
      {
        REP(y,k)s.pb('A');
        s.pb('B');
      }
      REP(x,p)s.pb('A');
      REP(x,q)s.pb('B');
      REP(x,j)
      {
        s.pb('A');
        REP(y,k)s.pb('B');
      }
      debug(s);
      cout<<s.substr(l-1,r-l+1)<<endl;
    }
}