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.*;
import java.math.BigInteger;

public class B1315{
    static int gcd(int a, int b) 
    { 
        if (a == 0) 
            return b; 
        if (b == 0) 
            return a; 
        int k; 
        for (k = 0; ((a | b) & 1) == 0; ++k)  
        { 
            a >>= 1; 
            b >>= 1; 
        } 
        while ((a & 1) == 0) 
            a >>= 1; 

        do { 
            while ((b & 1) == 0) 
                b >>= 1; 
            if (a > b)  
            { 
                int temp = a; 
                a = b; 
                b = temp; 
            } 

            b = (b - a); 
        } while (b != 0); 
        return a << k; 
    }   
    static int lcm(int a,int b){
        return a*b/gcd(a,b);
    }
	public static int binary(int arr[],int n, int a,int beg,int end){
		while(beg<=end){
			int mid=beg+(end-beg)/2;
			if(arr[mid]==a)
				return mid;
			if(arr[mid]<a)
				end=mid-1;
			else
				beg=mid+1;
		}
		return -1;
	}
    
    public static int binaryless(int arr[],int n, int a,int beg,int end){
        while(beg<=end){
            int mid=beg+(end-beg)/2;
            if(arr[mid]>=a){
                if(mid==1||arr[mid-1]<a)
                    return mid;
                end=mid-1;
            }
            else{
                beg=mid+1;
            }
        }
        return -1;
    }
	public static void sieve(int t){
		boolean srr[]=new boolean[t+1];
        
        for(int i=2;i*i<=t;i++)
        {
            if(!srr[i])
            {
                for(int k=i*i;k<=t;k+=i)
                    srr[k]=true;
            }
        }
        for(int i=2;i<t+1;i++){
        	if(!srr[i])
            	System.out.println(i);
        }
	}
    static int power(int x, int y, int p) 
    { 
        int res = 1;      
        x = x % p;  
        while (y > 0) 
        { 
            if((y & 1)==1) 
                res = (res * x) % p; 
            y = y >> 1;  
            x = (x * x) % p;  
        } 
        return res; 
    } 
    
    static class Custom implements Comparable<Custom>{
        int d;
        int f;
        public Custom(int d,int f){
            this.d=d;
            this.f=f;
        }
        public int compareTo(Custom t){
            return this.d-t.d;
        }
    }
    void solve() throws IOException {
        int t=nextInt();
        while(t--!=0){
            int arr[]=new int[2];
            arr[0]=nextInt();

            arr[1]=nextInt();
            int p=nextInt();
            char ch[]= next().toCharArray();
            int n=ch.length;
            int sum=0;
            sum+=arr[ch[n-2]-'A'];
            if(sum>p){
                pw.println(n);
                continue;
            }
            int i=0;
            for(i=n-3;i>=0;i--){
                if(ch[i]!=ch[i+1]){
                    sum+=arr[ch[i]-'A'];
                    if(sum>p){
                        i++;
                        break;
                    }
                }
            }
            i=Math.max(0,i);
            pw.println(i+1);
            // System.out.println("SDfs");
        }
    }
	String next() throws IOException {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(br.readLine());
        }
        return st.nextToken();
    }
    String nextLine()throws IOException{
        return br.readLine();
    }
    int nextInt() throws IOException {
        return Integer.parseInt(next());
    }
    long nextLong() throws IOException {
        return Long.parseLong(next());
    }
 
    StringTokenizer st;
    BufferedReader br;
    PrintWriter pw;
    Scanner sc;
    void run() throws IOException {
        sc=new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        br = new BufferedReader(new InputStreamReader(System.in));
        pw = new PrintWriter(System.out);
        long time = System.currentTimeMillis();
        solve();
        System.err.println("time = " + (System.currentTimeMillis() - time));
        pw.close();
    }
 
    public static void main(String[] args) throws IOException {
        new B1315().run();
    }
 
}