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 Homecoming{
    public static int next(char[] a, int index)
    {
        char b = a[index];
        while(true)
        {
            index++;
            if(index == a.length)
                return -1;
            if(b != a[index])
                return index;
                
        }
    }
     public static void main(String []args){
         Scanner s1 = new Scanner(System.in);
         int n = Integer.parseInt(s1.nextLine());
         for(int i = 0; i < n; i++)
         {
             String s = s1.nextLine();
             String[] splitStrings = s.split(" ");
             int a = Integer.parseInt(splitStrings[0]);
             int b = Integer.parseInt(splitStrings[1]);
             int p = Integer.parseInt(splitStrings[2]);
             s = s1.nextLine();
             s = s.trim();
             s = new StringBuilder(s).reverse().toString();
             s = " " + s;
             char[] c = s.toCharArray();
             
             int index = 2;
             while(true)
             {
                 if(c[index] == 'A')
                    p -= a;
                 else
                    p -= b;
                 if(p < 0)
                 {
                    System.out.println(c.length - index + 1);
                    break;
                 }
                 if(index == c.length - 1 && index == 2){
                    System.out.println(1);
                    break;
                 }
                 else
                 {
                     index = next(c, index);
                     if(index == -1)
                     {
                        System.out.println(1);
                        break;
                     }
                 }
             }
         }
     }
}