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.ArrayList;
import java.util.Collections;
import java.util.Scanner;

public class HomeComing {

	public static void main(String[] args) 
	{
		Scanner cin= new Scanner(System.in);
		int t=cin.nextInt();
		for (int i = 0; i < t; i++)
		{
			long a=cin.nextInt();
			long b=cin.nextInt();
			long p=cin.nextInt();
			StringBuilder str= new StringBuilder(cin.next());
			ArrayList<Integer> list= new ArrayList<>();
			ArrayList<Long> cost= new ArrayList<>();
			long totalcost=0;
			int j=0;
			int r=0;
			for (j = 0; j <str.length()-2; j++)
			{
				if(str.charAt((int) j)!=str.charAt((int) (j+1)))
				{
					list.add(j+2);
					if(str.charAt(0)=='B')
					{
						cost.add(b);
					}
					else
					{
						cost.add(a);
					}
					r++;
					break;
				}
			}
			for(j=j+1;j<=str.length()-2;j++)
			{
				if(str.charAt((int) j)!=str.charAt((int) (j+1)))
				{
					if(str.charAt((int) (j+1))=='B')
					{
						cost.add(a+cost.get(r-1));
						
					}
					else
					{
						cost.add(b+cost.get(r-1));
					}
					list.add(j+2);
					r++;
				}
			}
			if(r!=0&&list.get(r-1)!=str.length())
			{
				list.add(str.length());
				if(str.charAt(list.get(r-1))=='B')
				{
						cost.add(b+cost.get(r-1));
				}
				else
				{
						cost.add(a+cost.get(r-1));
				}
			}
			if(r==0)
			{
				if(str.charAt(r)=='B')
				{
						cost.add(b);
				}
				else
				{
						cost.add(a);
				
				}
				list.add(str.length());
				
			
			}
			totalcost=cost.get(list.size()-1);
		/*for (int k = 0; k < list.size(); k++) 
			{
				System.out.println(cost.get(k)+" " +list.get(k));
			}
		*/
			for (int k = 0; k <list.size(); k++) 
			{
				cost.set(k,cost.get(list.size()-1)-cost.get(k));
			}
		/*	for (int k = 0; k < list.size(); k++) 
			{
				System.out.println(cost.get(k)+" " +list.get(k));
			}
			
		*/	int flag1=0;
			if(p<a&&p<b)
			{
				System.out.println(str.length());
				continue;
			}
			if(p>=totalcost)
			{
				System.out.println(1);
			}
			else
			{
				for (int k = 0; k < list.size(); k++)
				{
					if(p>=cost.get(k))
					{
						System.out.println(list.get(k));
						flag1=1;
						break;
					}
				}
				if(flag1==0)
				{
					System.out.println(str.length());
				}
			}
		}
		

	}

}
