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.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.Writer;
import java.math.BigInteger;
import java.util.Scanner;
import java.util.StringTokenizer;
 
public class temp {
		
	void solve() throws IOException {
		
		FastReader sc = new FastReader();
		int t = sc.nextInt();
		while(t-->0)
		{
			int a = sc.nextInt();
			int b = sc.nextInt();
			int p = sc.nextInt();
			String s = sc.next();
			int n = s.length();
			
			if(n==1)
			{
				System.out.println(s.charAt(0)=='A' ? a : b);
				continue;
			}
			
			int A[] = new int[n];
			int B[] = new int[n];
			
			for(int i=0;i<n;i++)
			{
				if(i==0)
				{
					A[i] = (s.charAt(i)=='A' ? 1 : 0);
					B[i] = (s.charAt(i)=='B' ? 1 : 0);
				}
				else
				{
					if(s.charAt(i)!=s.charAt(i-1))
					{
						if(s.charAt(i)=='A')
						{
							A[i] = A[i-1]+1;
							B[i] = B[i-1];
						}
						else
						{
							A[i] = A[i-1];
							B[i] = B[i-1]+1;
						}
					}
					else
					{
						A[i] = A[i-1];
						B[i] = B[i-1];
					}
				}
			}
			
			
			int idx = n-1;
			for(int i=0;i<n;i++)
			{
				int aa = A[n-1] - A[i];
				int bb = B[n-1] - B[i];
				
				if(s.charAt(i)=='A')
					aa++;
				else
					bb++;
				
				if(s.charAt(n-1)!=s.charAt(n-2))
				{
					if(s.charAt(n-1)=='B')
						bb--;
					else
						aa--;
				}
				long pay = (long)aa*(long)a + (long)bb*(long)b;
				if(pay <= p)
				{
					idx = i;
					break;
				}
			}
			System.out.println(idx+1);
		}
    }
	
	public static void main(String[] args) throws IOException {
		// TODO Auto-generated method stub
		new temp().solve();
	}
	
	class FastReader {
		BufferedReader br;
		StringTokenizer st;
 
		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
 
		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
 
		int nextInt() {
			return Integer.parseInt(next());
		}
 
		long nextLong() {
			return Long.parseLong(next());
		}
 
		double nextDouble() {
			return Double.parseDouble(next());
		}
 
		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}
}