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.*;
public class dkbproblem {
	public static void main(String[] args) {
		Scanner dek = new Scanner(System.in);
		int t = dek.nextInt();
		while (t-- > 0) {
for(int k=0;k<2;k++){
	
}
			int a = dek.nextInt();
			int b = dek.nextInt();
			int p = dek.nextInt();
			ArrayList<pair> arrl = new ArrayList<>();
			String stri = dek.next();

			int ival = 0;

			while (ival < stri.length() - 1) {

				char ch = stri.charAt(ival);
				int c1 = 0;
				if (ch == 'A') {
					while (ival < stri.length() - 1 && stri.charAt(ival) == 'A') {
						c1++;
						ival++;
					}
				} else {
					while (ival < stri.length() - 1 &&stri.charAt(ival) == 'B') {
						c1++;
						ival++;
					}
				}
				for(int k=0;k<2;k++){
					
				}
				arrl.add(new pair(c1, ch));
			}
			ival = arrl.size() - 1;
			int anss = stri.length();
			while (ival >= 0) {

				if (arrl.get(ival).c == 'A') {
					if (a > p) {
						break;
					}
					p -= a;
					anss -= arrl.get(ival).frqq;
				} else {
					if (b > p) {
						break;
					}
					anss -= arrl.get(ival).frqq;
					p -= b;
				}
				ival--;
				if (p <= 0) {
					break;
				}

			}
			for(int k=0;k<2;k++){
				
			}
			System.out.println(anss);

		}
	}

	private static class pair {
		char c;
		int frqq;

		pair(int f, char c) {
			this.frqq = f;
			this.c = c;
		}
	}

}