After a long party Petya decided to return home, but he turned out to be at the opposite end of the town from his home. There are n crossroads in the line in the town, and there is either the bus or the tram station at each crossroad.

The crossroads are represented as a string s of length n, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad. Currently Petya is at the first crossroad (which corresponds to s_1) and his goal is to get to the last crossroad (which corresponds to s_n).

If for two crossroads i and j for all crossroads i, i+1, …, j-1 there is a bus station, one can pay a roubles for the bus ticket, and go from i-th crossroad to the j-th crossroad by the bus (it is not necessary to have a bus station at the j-th crossroad). Formally, paying a roubles Petya can go from i to j if s_t = A for all i ≤ t < j. 

If for two crossroads i and j for all crossroads i, i+1, …, j-1 there is a tram station, one can pay b roubles for the tram ticket, and go from i-th crossroad to the j-th crossroad by the tram (it is not necessary to have a tram station at the j-th crossroad). Formally, paying b roubles Petya can go from i to j if s_t = B for all i ≤ t < j.

For example, if s="AABBBAB", a=4 and b=3 then Petya needs:

<image>

  * buy one bus ticket to get from 1 to 3, 
  * buy one tram ticket to get from 3 to 6, 
  * buy one bus ticket to get from 6 to 7. 



Thus, in total he needs to spend 4+3+4=11 roubles. Please note that the type of the stop at the last crossroad (i.e. the character s_n) does not affect the final expense.

Now Petya is at the first crossroad, and he wants to get to the n-th crossroad. After the party he has left with p roubles. He's decided to go to some station on foot, and then go to home using only public transport.

Help him to choose the closest crossroad i to go on foot the first, so he has enough money to get from the i-th crossroad to the n-th, using only tram and bus tickets.

Input

Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4).

The first line of each test case consists of three integers a, b, p (1 ≤ a, b, p ≤ 10^5) — the cost of bus ticket, the cost of tram ticket and the amount of money Petya has.

The second line of each test case consists of one string s, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad (2 ≤ |s| ≤ 10^5).

It is guaranteed, that the sum of the length of strings s by all test cases in one test doesn't exceed 10^5.

Output

For each test case print one number — the minimal index i of a crossroad Petya should go on foot. The rest of the path (i.e. from i to n he should use public transport).

Example

Input


5
2 2 1
BB
1 1 1
AB
3 2 8
AABBBBAABB
5 3 4
BBBBB
2 1 1
ABABAB


Output


2
1
3
1
6
import java.util.ArrayList;
import java.util.Scanner;

public class main {
    
    
    public static void main(String[] args){
        
        Scanner in = new Scanner(System.in);
        
        int t = in.nextInt();
        
        m: for(int j=0;j<t;j++){
        
        int a = in.nextInt();
        
        int b = in.nextInt();
        
        int c = in.nextInt();
        
        String s = in.next();
        
        ArrayList<String> list = new ArrayList();
        ArrayList<Integer> taka = new ArrayList();
        
        if(a>c && b>c){
            System.out.println(s.length());
            continue m;
        }
        else{
            int count=1;
            
            char x[] = s.toCharArray();
            char kabadi=x[0]; int suru=0;
            for(int i=0;i<x.length;i++){
                if(kabadi!=x[i]){
                    if(x[i]=='A'){
                        list.add(suru+1 +"-"+(i+1));
                        suru=i;
                        taka.add(b);
                        kabadi=x[i];
                    }
                    else if(x[i]=='B'){
                        list.add((suru+1) +"-"+(i+1));
                        suru=i;
                        taka.add(a);
                        kabadi=x[i];
                    }
                }
            }
            int lastIndex=0;
            try{
            lastIndex=Integer.parseInt(list.get(list.size()-1).split("-")[1]);}
            catch(Exception e){
                lastIndex=1;
            }
            if(lastIndex!=s.length()){
                if(x[lastIndex-1]=='A'){
                    taka.add(a);
                    list.add(lastIndex +"-"+s.length());
                }
                else if(x[lastIndex-1]=='B'){
                    taka.add(b);
                    list.add(lastIndex +"-"+s.length());
                }
            }
            
        }
     //   System.out.println(taka);
      //  System.out.println(list);
        int cost=0;
        int hataSuru=0;
        for(int i=taka.size()-1;i>=0;i--){
            cost+=taka.get(i);
            if(c<cost){
                hataSuru=i;
                break;
            }
        }
        
        if(c-cost >= 0){
            System.out.println(1);
        }
        
        else
        System.out.println(Integer.parseInt(list.get(hataSuru).split("-")[1]));
        
        }
    }
    
}
