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;

import java.util.ArrayList;

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

		int t = sc.nextInt();

		while (t > 0)
		{			
			int a = sc.nextInt();

			int b = sc.nextInt();

			int p = sc.nextInt();

			ArrayList<Integer> list = new ArrayList<>();

			StringBuilder sb = new StringBuilder();

			String str1 = sc.next();
	
			int i = 0;

			int s = 0;
				
			int len1 = str1.length();

			if (len1 == 1)
			{
				System.out.println(1);

				t --;

				continue;
			}
			
			while (i < len1) {

				int cnt = 0;
            
				char c = str1.charAt(i);
				
				sb.append(c);
				
				i ++;

				cnt ++;
				
				while (i < len1 && str1.charAt(i) == c) {
				
					i ++;

					cnt ++;
				}

				list.add(cnt);

				s += cnt;
			}

			if (a > p && b > p)
			{
				System.out.println(s);

				t --;

				continue;
			}

			String str2 = sb.toString();

			int sum = 0;
			
			int len2 = str2.length();

			int j = len2 - 1;

			if (len2 == 1)
			{
				char c = str2.charAt(j);
				
				if (c == 'A')
				{
					if (a <= p)
					{
						System.out.println(1);
					}else{
						System.out.println(s);
					}
				}else{
					if (b <= p)
					{
						System.out.println(1);
					}else{
						System.out.println(s);
					}
				}

				t --;

				continue;
			}

			char ch = str2.charAt(j);

			if (ch == 'A' && a > p && (list.get(j) != 1))
			{
				System.out.println(s);

				t --;

				continue;
			}else if (ch == 'B' && b > p && (list.get(j) != 1)){
				System.out.println(s);

				t --;

				continue;
			}
		
			if (list.get(j) == 1)
			{
				j --;

				while (j >= 0)
				{
					char c = str2.charAt(j);

					if (c == 'A')
					{
						if (p >= a)
						{
							p -= a;

							sum += list.get(j);
						}else{
							break;
						}
					}else{
						if (p >= b)
						{
							p -= b;
							
							sum += list.get(j);
						}else{
							break;
						}
					}
					
					j --;
				}

				System.out.println(s - sum);
			}else{
				while (j >= 0)
				{
					char c = str2.charAt(j);

					if (c == 'A')
					{
						if ((p -= a) >= 0)
						{
							sum += list.get(j);
						}else{
							break;
						}
					}else{
						if ((p -= b) >= 0)
						{
							sum += list.get(j);
						}else{
							break;
						}
					}
					
					j --;
				}

				System.out.println(s - sum + 1);
			}

			t --;
		}	
	}
}