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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;


public class Main{
   public static void main(String[] args) {
	  FastReader sc = new FastReader();
	
	  int tests = sc.nextInt();
	  
	  while(tests > 0) {
		 tests--;
		 int a = sc.nextInt();
		 int b = sc.nextInt();
		 int p = sc.nextInt();
		 
		 String s = sc.next();
		 
		 int len = s.length();
		 int left = 0;
		 int right = len-1;
		 
		 while(left < right) {
			 int mid = left + (right - left) / 2;
			 
			 if(afford(mid, s, p, a, b)) {
				 right = mid;
			 }
			 
			 else {
				 left = mid + 1;
			 }
		}
		 
		 System.out.println(left + 1);
	}
	  
}
   
   public static boolean afford(int index, String s, int p, int a, int b) {
	   
	   while(p >= 0 && index < s.length()) {
		   if(index == s.length() - 1) break;
		   
		   
		   if(s.charAt(index) == 'A') {
			   while(index < s.length()&& s.charAt(index) == 'A') {
				   index++; 
			   }
			   p -= a;
		   }
		   
		   else {
			   while(index < s.length()&& s.charAt(index) == 'B') {
				   index++;
			   }
			   p-=b;
		   }
		   
	   }
	   return p >= 0;
   }
 
   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;
	    }
	}
     
}