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.*;
import java.math.*;

public class B {

	public void realMain() throws Exception {

		BufferedReader fin = new BufferedReader(new InputStreamReader(System.in), 1000000);

		String in = fin.readLine();

		String[] ar = in.split(" ");

		int T = Integer.parseInt(ar[0]);
	for(int t = 0; t < T; t++) {
		in = fin.readLine();
		ar = in.split(" ");

		int a = Integer.parseInt(ar[0]);
		int b = Integer.parseInt(ar[1]);
		int p = Integer.parseInt(ar[2]);

		in = fin.readLine();
		char[] c = in.toCharArray();

		int n = in.length();

		/*char[] c = new char[n];

		for(int i = 0; i < n - 1; i++) {
			c[i] = in.charAt(i);
		}

		if(c[n - 2] == 'A') {
			c[n - 1] = 'B';
		} else {
			c[n - 1] = 'A';
		}
*/

		int canmakeind = n - 1;

		long curcost = 0;

		char tmpcur = c[n - 1];

		int numatstart = 0;

		int start = -1;
		for(int j = n - 1; j >= 0; j--) {
			if(c[j] != tmpcur) {
				start = j;
				break;
			}
			numatstart++;
		}

		
		if(start == -1) {
			if(tmpcur == 'A') {
				if(a <= p) {
					System.out.println(1);
					continue;
				} else {
					System.out.println(n);
					continue;
				}
			} else {
				if(b <= p) {
					System.out.println(1);
					continue;
				} else {
					System.out.println(n);
					continue;
				}
			}
		}

		if(numatstart != 1) {
			if(c[n - 1] == 'A') {
				curcost += a;
			} else {
				curcost += b;
			}

			if(curcost <= p) {
				canmakeind = start + 1;
			} else {
				System.out.println(canmakeind + 1);
				continue;
			}
		}



		for(int j = start; j >= 0; j--) {
			char cur = c[j];

			int diffind = -1;

			for(int i = j - 1; i >= 0; i--) {
				if(c[i] != cur) {
					diffind = i;
					break;
				}
			}

			if(diffind == -1) {
				if(cur == 'A') {
					curcost += a;
				} else {
					curcost += b;
				}
				j = -1;

				if(curcost <= p) {
					canmakeind = 0;
				}
			} else {
				if(cur == 'A') {
					curcost += a;
				} else {
					curcost += b;
				}
				
				j = diffind + 1;
				
				if(curcost <= p) {
					canmakeind = diffind + 1;
					//System.err.println("did it " + cur + diffind);
				} else {
					break;
				}

				
			
			}

			
		}

		if(curcost <= p) {
			canmakeind = 0;
		}

		System.out.println(canmakeind + 1);

	}


	}


	public static void main(String[] args) throws Exception {
		B b = new B();
		b.realMain();
	}
}