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.*;
import java.io.*;

public class B {
	public static void main(String[] args) {
		InputReader sc = new InputReader(System.in);
		int t = sc.nextInt();
		while (t-- > 0) {
			int a = sc.nextInt();
			int b = sc.nextInt();
			int p = sc.nextInt();
			String str = sc.next();
			int arr[] = new int[str.length()];
			//arr[str.length()-1] = str.charAt(str.length()-1) == 'A' ? a : b;
			int res = str.length()-1;
			for (int i = str.length() - 2 ; i >= 0 && arr[i+1] <= p ; i--) {
				if (str.charAt(i) != str.charAt(i+1)) {
					if (i == str.length() - 2) {
						arr[i] = str.charAt(i) == 'A' ? a : b;
					}else {
						arr[i] = (str.charAt(i) == 'A' ? a : b ) + arr[i+1];
					}
				}else {
					if (i == str.length() - 2) {
						arr[i] = str.charAt(i) == 'A' ? a : b;
					}else {
						arr[i] = arr[i+1];
					}
				}
				if (arr[i] > p) {
					break;
				}
				res = i;
			}
			//System.out.println(a+" "+b+" "+p);
			//System.out.println(Arrays.toString(arr));
			System.out.println(res+1);
		}
	}
	
	static class InputReader{
		BufferedReader br;
		StringTokenizer st;
        
        public InputReader(InputStream in){
        	try {
        		br = new BufferedReader(new
        				InputStreamReader(in)); 
        	}catch (Exception e) {
        		System.out.println(e.toString());
        	}
        }
        
        String next(){
        	while (st == null || !st.hasMoreElements()){
        		try{
        			st = new StringTokenizer(br.readLine()); 
                }catch (IOException  e){
                	e.printStackTrace(); 
                }
            }
            return st.nextToken(); 
        }
        
        int nextInt(){
        	return Integer.parseInt(next()); 
        }
        
        long nextLong(){
        	return Long.parseLong(next()); 
        }
        
        double nextDouble(){
        	return Double.parseDouble(next()); 
        }
        
        String nextLine(){
        	String str = "";
        	try{
        		str = br.readLine();
        	}catch (IOException e){
        		e.printStackTrace(); 
            }
            return str; 
        }
    }
}
