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.*;
import java.util.*;
 
public class codeforces {
	public static PrintWriter out;
	public static class MyScanner {
		BufferedReader br;
		StringTokenizer st;
		public MyScanner() {
			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;
		}
	}
	// The next two void functions are shuffle functions on an array.
	// This way, quick sort encounters the worst case scenario with low probability.
	// A reason for doing this is if we don't a primitive data array to an object array.
	// For safety reasons, it would be preferable to use object arrays unless doing so expends a lot of time.
	public static void shuffleLongArray(long[] arr) {
		int n = arr.length;
		Random rnd = new Random();
        for (int i=0; i < n; ++i) {
        	long tmp = arr[i];
        	int randomPos = i + rnd.nextInt(n-i);
        	arr[i] = arr[randomPos];
        	arr[randomPos] = tmp;
        }
	}
	public static void shuffleIntArray(int[] arr) {
		int n = arr.length;
		Random rnd = new Random();
        for (int i=0; i < n; ++i) {
        	int tmp = arr[i];
        	int randomPos = i + rnd.nextInt(n-i);
        	arr[i] = arr[randomPos];
        	arr[randomPos] = tmp;
        }
	}
	public static void main (String[] args) throws IOException {
    	MyScanner sc = new MyScanner();
    	out = new PrintWriter(new BufferedOutputStream(System.out));
    	// Write your solution here.
    	int t = sc.nextInt();
    	for (int asap=0; asap < t; ++asap) {
    		long a = sc.nextLong(); long b = sc.nextLong(); long p = sc.nextLong();
    		String s = sc.next();
    		long[] dp = new long[s.length()]; // price
    		s = s.substring(0, s.length()-1) + "C"; // for calculations
    		// out.println(s);
    		int offa = s.length()-1; int offb = s.length()-1;
    		for (int i=s.length()-2; i >= 0; --i) {
    			if (s.charAt(i)=='A') {
    				if (s.charAt(i+1)!='A') offa = i+1;
    				dp[i] = dp[offa] + a;
    			}
    			else if (s.charAt(i)=='B') {
    				if (s.charAt(i+1)!='B') offb = i+1;
    				dp[i] = dp[offb] + b;
    			}
    		}
    		/*
    		for (int i=0; i < s.length(); ++i) {
    			out.print(dp[i] + " ");
    		}
    		*/
    		// out.println();
    		for (int i=0; i < s.length(); ++i) {
    			if (dp[i] <= p) {
    				out.println(i+1);
    				break;
    			}
    		}
    	}
    	out.close();
	}
}