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
//package codeforces;
import java.io.PrintWriter;
import java.util.*;
public class solution { 
    public static void main(String args[]){ 
		Scanner s=new Scanner(System.in);  
		PrintWriter out=new PrintWriter(System.out);
        int t=s.nextInt();
        for(int tt=0;tt<t;tt++) {
        	int a=s.nextInt(),b=s.nextInt();long p=s.nextInt();
        	String d=s.next();
        	ArrayList<Integer> e=new ArrayList<Integer>();
        	ArrayList<Integer> f=new ArrayList<Integer>();
        	long ans=0;
        	int count=1;
        	for(int i=1;i<d.length();i++) {
        		if(d.charAt(i)==d.charAt(i-1)) {
        			count++;
        		}
        		else {
        			if(d.charAt(i-1)=='A') {
        				e.add(a);
        				ans+=a;
        			}else {
        				e.add(b);
        				ans+=b;
        			}
        			f.add(i+1);
        			count=1;
        		}
        	}
        	if(count!=1) {
        		if(d.charAt(d.length()-1)=='A') {
    				e.add(a);
    				ans+=a;
    			}else {
    				e.add(b);
    				ans+=b;
    			}
    			f.add(d.length());
    			count=1;
        	}
        	int l=1;
        	int i=0;
        	while(ans>p && f.size()!=i) {
        		l=f.get(i);
        		ans-=e.get(i);
        		i++;
        	}
        	if(ans>p) {
        		ans=d.length();
        	}
        	System.out.println(l);
        }
        out.close();
	    s.close();
    } 
    
    static void sort(int [] a) {
		ArrayList<Integer> l=new ArrayList<>();
		for (int i:a) l.add(i);
		Collections.sort(l);
		for (int i=0; i<a.length; i++) a[i]=l.get(i);
	}
    static int gcd(int a, int b)
    {
      if (b == 0)
        return a;
      return gcd(b, a % b);
    }
    static void sortcol(int a[][],int c) {
		Arrays.sort(a, (x, y) -> {
			if (x[c] != y[c]) return(int)( x[c] - y[c]);
			return (int)-(x[1]+x[2] - y[1]-y[2]);
		});
	}
    
}
