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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.Scanner;
public class CodeForce {
	public static void main(String[] args) throws IOException {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
                
                   int t=Integer.parseInt(br.readLine());
                   while(t-->0){
                       String[] s=br.readLine().split(" ");
                       int a=Integer.parseInt(s[0]);
                       int b=Integer.parseInt(s[1]);
                       int p=Integer.parseInt(s[2]);
                       String str=br.readLine();
                       int flag=0;
                       char ch=str.charAt(0);
                       int c=0;
                       int a1=0,b1=0;
                       int[][] arr=new int[str.length()][2];
                       for(int i=1;i<str.length();i++){
                           if(ch!=str.charAt(i)){
                               if(ch=='B'){ 
                                   arr[c][0]=2;
                                   arr[c][1]=i;
                                   c++;
                                   a1++;
                               }
                                   
                               else{
                                   arr[c][0]=1;
                                   arr[c][1]=i;
                                   c++;
                                   b1++;
                               }
                                  
                               ch=str.charAt(i);
                           }
                          
                       }
                       
//                       for(int i=0;i<c;i++)
//                           System.out.println(arr[i][0]+" "+arr[i][1]);
                       if(c==0){
                           if(str.charAt(0)=='B'){
                               if(p<b)
                                   System.out.println(str.length());
                               else
                                   System.out.println(1);
                           }
                           else{
                               if(p<a)
                                   System.out.println(str.length());
                               else
                                   System.out.println(1);
                           }
                           flag=1;
                       }
                      
                       else if(str.charAt(str.length()-1)==str.charAt(str.length()-2)){
                           if(arr[c-1][0]==1){
                               arr[c][0]=2;
                               arr[c][1]=str.length()-1;
                           }
                           else{
                                arr[c][0]=1;
                               arr[c][1]=str.length()-1;
                           }
                           c++;
                       }
                      
                       for(int i=c-1;i>=0;i--){
                           if(arr[i][0]==1){
                               if(p<a){
                                   System.out.println(arr[i][1]+1);
                                  // System.out.println(i+" "+arr[i][1]);
                                  flag=1;
                                   break;
                               }
                               p=p-a;
                           }
                           else{
                               if(p<b){
                                   System.out.println(arr[i][1]+1);
                                   //System.out.println(i+" "+arr[i][1]);
                                   flag=1;
                                   break;
                               }
                               p=p-b;
                           }
                       }
                       if(flag==0)
                           System.out.println(1);
                           
                   }
                    
                    
                    
                }
		

}