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.*;
public class Codeforces{
	public static int get(List<Long> vals,long key) {
		int lo=0;
		int hi=vals.size()-1;
		int x=rank(vals,lo,hi,key);
		return x;
	}
	private static int rank(List<Long> vals,int lo,int hi,long key){
		while(lo<=hi) {
			int mid=lo+(hi-lo)/2;
			if(key<vals.get(mid)) hi=mid-1;
			else if(key>vals.get(mid)) lo=mid+1;
			else if(key==vals.get(mid)) return mid;
		}
		return lo;
	}
	
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		for(int i=0;i<t;i++) {
			int a=sc.nextInt();
			int b=sc.nextInt();
			int p=sc.nextInt();
			sc.nextLine();
			String s=sc.nextLine();
			List<Integer> index=new LinkedList<>();
			List<Long> cost=new LinkedList<>();
			long total_cost=0;
			boolean temp=true;
			
			if(String.valueOf(s.charAt(0)).equals("A")) {
				temp=true;
			}
			else temp=false;
			index.add(1);
			cost.add(Long.valueOf(0));
			
			for(int j=1;j<s.length()-1;j++) {
				if((temp==true && String.valueOf(s.charAt(j)).equals("A")) || (temp==false && String.valueOf(s.charAt(j)).equals("B"))) {
					continue;
				}
				else {
					if(temp) {
						total_cost+=a;
						cost.add(total_cost);
						temp=!temp;
					}
					else {
						total_cost+=b;
						cost.add(total_cost);
						temp=!temp;
					}
					index.add(j+1);
				}
			}
			if(temp) total_cost+=a;
			else total_cost+=b;
			cost.add(total_cost);
			index.add(s.length());
			
			int ind=get(cost,total_cost-p);
			System.out.println(index.get(ind));
		}
	}
}