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 javax.management.DynamicMBean;
import javax.print.DocFlavor;
import javax.print.attribute.HashAttributeSet;
import java.awt.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.nio.file.attribute.FileAttribute;
import java.time.DayOfWeek;
import java.time.LocalDate;
import java.time.Month;
import java.time.temporal.TemporalAdjusters;
import java.util.List;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.*;
import java.util.Arrays;
import java.util.concurrent.Callable;
import java.time.DayOfWeek;
import java.time.LocalDate;
import java.time.Month;
import java.time.temporal.TemporalAdjusters;



public class Main {


    static 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;
        }
    }

    public static void main(String[] args) {
        FastReader sc = new FastReader();

        int t=sc.nextInt();
        for(int i=0;i<t;i++){
            int a=sc.nextInt();
            int b=sc.nextInt();
           int p=sc.nextInt();
           String s=sc.next();
           int[] dp1=new int[s.length()];
           int[] dp2=new int[s.length()];
           char ch=s.charAt(0);
           long sum=0;
                   int index=0;
           if(p<a && p<b){
               System.out.println(s.length());
               continue;
           }
           for(int j=1;j<s.length();j++){
               if(s.charAt(j)==ch){
                   if(j==s.length()-1){
                       dp1[index]=ch=='A'?a:b;
                       dp2[index]=j+1;
                       sum+=dp1[index];
                       index++;
                   }
                   continue;
               }
               else{
                   dp1[index]=ch=='A'?a:b;
                   dp2[index]=j+1;
                   sum+=dp1[index];
                   index++;
                   ch=s.charAt(j);
               }

           }
           if(p>=sum){
               System.out.println(1);
               continue;
           }
           int pos=0;
           for(int j=0;j<index;j++){
               sum-=dp1[j];
               if(p>=sum){
                   pos=dp2[j];
                   break;
               }
           }
           System.out.println(pos);

        }
    }
}



