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.lang.Math;
import java.io.* ;

public class Account {
	
	class Pair {
        int key ;
        int val ;
        
        Pair(int key ,int val) {
            this.key = key ;
            this.val = val ;
        }
        
        int getKey() {
        	return this.key ;
        }
        
        int getVal() {
        	return this.val ;
        }
    }
	
	public static void main (String[] args) {
		
//		Scanner sc= new Scanner(System.in) ;
		FastReader sc= new FastReader() ;
//		StringBuilder res= new StringBuilder() ;
		int t= 1 ;  
		t= sc.nextInt() ;
		while(t-- > 0) {
			int a = sc.nextInt() ;
			int b = sc.nextInt() ;
			int p = sc.nextInt() ;
			String str = sc.next() ;
			char[] arr = str.toCharArray() ;
			ArrayList<Integer> lst = new ArrayList<>() ;
			long sum= 0 ;
			int n= arr.length ,in= n-1 ;
			lst.add(0) ;
//			AABBAAA
			for(int i=1 ;i<n ;i++) {
				if(arr[i] != arr[i-1]) {
					lst.add(i) ;
					if(arr[i-1] == 'A') {
						sum += a ;
					}else {
						sum += b ;
					}
				}
			}if(lst.get(lst.size()-1) != n-1) {
				if(arr[n-1] == 'A') {
					sum += a ;
				}else {
					sum += b ;
				}
				lst.add(n-1) ;
			}
//			System.out.println(sum+" "+lst);
			for(int i=0 ;i<lst.size() ;i++) {
//				System.out.println(sum+" "+i);
				if(sum <= p) {
					in= lst.get(i) ;
					break ;
				}else {
					if(arr[lst.get(i)] == 'A') {
						sum -= a ;
					}else {
						sum -= b ;
					}
				}
			}
			System.out.println(in+1);
//			System.out.println(sum+" "+lst+" "+in);
		}
	}
	

	public static int gcd(int a, int b)
	{
	    if (b == 0)
	        return a;
	    return gcd(b, a % b);
	     
	}

	
	public static class FastReader {
        
		BufferedReader br;
        StringTokenizer st;
 
        public FastReader()
        {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }
 
        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;
        }
 
        int [] readintarray(int n) {
            int res [] = new int [n];
            for(int i = 0; i<n; i++)res[i] = nextInt();
            return res;
        }
        long [] readlongarray(int n) {
            long res [] = new long [n];
            for(int i = 0; i<n; i++)res[i] = nextLong();
            return res;
        }
    }	
	
}