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.util.regex.*;
    import java.lang.*;
    import java.io.*;
     
    public class rough{

        public static long calc(char[] arr, int[] rem, long[] cost, int a, int b, int idx){
            if (idx >= arr.length)
                return 0;

            if (arr[idx-1] == 'A'){
                cost[idx] = a + calc(arr, rem, cost, a, b, rem[idx]);
            }else{
                cost[idx] = b + calc(arr, rem, cost, a, b, rem[idx]);
            }

            return cost[idx];
        }
     
        public static void main(String args[]) throws IOException {
            
            BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
            BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));

            int t = Integer.parseInt(br.readLine().trim());

            while (t-->0){

                String[] tmp = br.readLine().trim().split(" ");
                int a = Integer.parseInt(tmp[0]);
                int b = Integer.parseInt(tmp[1]);
                int p = Integer.parseInt(tmp[2]);

                char[] arr = br.readLine().trim().toCharArray();
                int[] rem = new int[arr.length+1];

                rem[arr.length] = arr.length;

                for (int i=arr.length-1 ; i>0 ; i--){
                    if (arr[i-1] == arr[i]){
                        rem[i] = rem[i+1];
                    }else{
                        rem[i] = i+1;
                    }
                }

                long[] cost = new long[arr.length+1];

                calc(arr, rem, cost, a, b, 1);

                int i=1; 

                while (i<arr.length && p < cost[i]){
                    i = rem[i];
                }

                System.out.println(i);
                //bw.write((i+1)+"\n");
            }
        }
    }