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.*;
import java.io.*;

public class B_1315 {
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		
		int t = sc.nextInt();
		while(t-->0) {
			int a = sc.nextInt(), b = sc.nextInt(), p = sc.nextInt();
			String s = sc.next();
			int n = s.length();
			ArrayList<Integer> count = new ArrayList<>(), type = new ArrayList<>();
			int c = 1;
			for(int i = 1; i < n; i++) {
				if(s.charAt(i) != s.charAt(i - 1)) {
					count.add(c);
					type.add(s.charAt(i - 1) == 'A' ? 1 : 2);
					c = 0;
				}
				c++;
			}
			count.add(c);
			type.add(s.charAt(n - 1) == 'A' ? 1 : 2);
			long cnt = 0;
			for(int i = 0; i < type.size() - 1; i++)
				cnt += type.get(i) == 1 ? a : b;
			if(count.get(count.size() - 1) > 1) {
				cnt += type.get(count.size() - 1) == 1 ? a : b;
				count.set(count.size() - 1, count.get(count.size() - 1) - 1);
				count.add(1);
				type.add(1);
			}
			
			int ans = 1;
			for(int i = 0; i < count.size() - 1; i++) {
				if(cnt <= p)
					break;
				cnt -= type.get(i) == 1 ? a : b;
				ans += count.get(i);
			}
			
			pw.println(ans);
		}
		pw.flush();
	}

	public static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public int[] nextIntArray(int n) throws IOException {
			int[] array = new int[n];
			for (int i = 0; i < n; i++)
				array[i] = nextInt();
			return array;
		}

		public Integer[] nextIntegerArray(int n) throws IOException {
			Integer[] array = new Integer[n];
			for (int i = 0; i < n; i++)
				array[i] = new Integer(nextInt());
			return array;
		}

		public long[] nextLongArray(int n) throws IOException {
			long[] array = new long[n];
			for (int i = 0; i < n; i++)
				array[i] = nextLong();
			return array;
		}

		public double[] nextDoubleArray(int n) throws IOException {
			double[] array = new double[n];
			for (int i = 0; i < n; i++)
				array[i] = nextDouble();
			return array;
		}

		public static int[] shuffle(int[] a) {
			int n = a.length;
			Random rand = new Random();
			for (int i = 0; i < n; i++) {
				int tmpIdx = rand.nextInt(n);
				int tmp = a[i];
				a[i] = a[tmpIdx];
				a[tmpIdx] = tmp;
			}
			return a;
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
}
