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

//package codeforces_contests;

import java.util.Scanner;

/**
 *
 * @author milon
 */
public class Contest1315problemB {
    public static void main(String[] args) {
        Scanner sin = new Scanner(System.in);
        int t=sin.nextInt();
        for (int ii = 0; ii < t; ii++) {
            long bus=sin.nextLong();
            long trump=sin.nextLong();
            long money=sin.nextLong();
            //AA BBBB AA BB
            //AABBBBAABB
            
            String root=sin.next();
            for(int i =root.length()-1; i >=0 ; i--) {//start form last
                //if(i==0){
                if(i==0 ){
                    System.out.println("1");
                    break;
                }
                //BBB(B)
//                if((i==(root.length()-1)) ){
//                    System.out.println("ok=1");
//                    break;
//                }
                
               
                //System.out.println(i+"=we are in station: "+root.charAt(i)+" currentm="+money);
                //System.out.println("we need $ tk to come here");
                char cur_position=root.charAt(i);//A
                char pre_position=root.charAt(i-1);//A
                
               
                    
                    //current cost
                    long cost=0;
                    if(pre_position=='A'){//cost=bus
                        cost=bus; 
                    }else{//cost=trump
                       cost=trump;
                    }
                    //money
                    //System.out.println("cost"+cost+" money="+money+" pre--"+ pre_position+" curr--"+cur_position);
                    if(pre_position==cur_position){//pre_position is same then contnue;
                        if((i==(root.length()-1))){
                            if(cost<=money){
                                money-=cost;
                            }else{
                                System.out.println(""+(1+i));
                                break;
                            }
                        }else{
//                            if(cost>money){
//                                System.out.println(""+(1+i));
//                                break;
//                            }
                        }
                    } 
                    else if(pre_position!=cur_position){
                        if(cost<=money){
                            money-=cost;
                        }else{
                            System.out.println(""+(1+i));
                            break;
                        }
                    }
                
            }
        }
    }
}
