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.Scanner;

public class Homecoming {

	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);

		int t, a, b, p;
		String s = "";
		t = scan.nextInt();
		for (int i = 0; i < t; i++) {
			a = scan.nextInt(); // cost of bus
			b = scan.nextInt(); // cost of tram
			p = scan.nextInt(); // amount of money

			// bus = A, tram = B
			s = scan.next();
			char[] c = new char[s.length()];
			c = s.toCharArray();

			char prev = 'c';
			int pay = 0;
			int take = 1;


				if (c[s.length()-1] != c[s.length()-2]) {
					if (c[s.length()-2] == 'A') {
						if (a <= p) {
							pay -= b;
						}
					} else {
						if (b <= p) {
							pay -= a;
						}
					}
				}
			
			for (int j = s.length() - 1; j >= 0; j--) {

				if (c[j] != prev) {
					prev = c[j];

					if (prev == 'A') {
						if (pay + a > p) {
							System.out.println((j + take));
							break;
						}
						pay += a;
					} else {
						if (pay + b > p) {
							System.out.println((j + take));
							break;
						}
						pay += b;
					}

				}
				if (j == 0) {
					System.out.println(1);
				}
				take = 2;
			}

		}
		scan.close();
	}
}
