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.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class TaskB {
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st;
        PrintWriter pw = new PrintWriter(System.out);
        
        int t = Integer.parseInt(br.readLine());
        while(t-->0) {
            st = new StringTokenizer(br.readLine());
            int a = Integer.parseInt(st.nextToken());
            int b = Integer.parseInt(st.nextToken());
            int p = Integer.parseInt(st.nextToken());

            char[] c = br.readLine().toCharArray();
            int n = c.length;

            long[] pay = new long[n];

            long total = 0;

            for(int i=0 ; i<n-1 ; i++) {
                if(i==0) {
                    if(c[i] == 'B') {
                        pay[i] = b;
                    }
                    else {
                        pay[i] = a;
                    }
                    total += pay[i];
                }
                else {
                    if(c[i] == c[i-1]) {
                        pay[i] = pay[i-1];
                    }
                    else {
                        if(c[i] == 'B') {
                            pay[i] = b;
                        }
                        else {
                            pay[i] = a;
                        }
                        total += pay[i];
                    }
                }
            }

            if(p < a && p < b) {
                pw.println(n);
            }
            else if(p >= total) {
                pw.println(1);
            }
            else {
                int ind = -1;
                for(int i=0 ; i<n-1 ; i++) {
                    if(i==0) {
                        total -= pay[i];
                    }
                    else {
                        if(c[i] != c[i-1]) {
                            total -= pay[i];
                        }
                    }
                    if(p >= total) {
                        ind = i;
                        break;
                    }
                }

                if(ind == -1) {
                    pw.println(n);
                }
                else {
                    while(ind < n-1 && c[ind] == c[ind+1]) {
                        ind++;
                    }
                    if(ind == n-1 || ind == n-2) {
                        pw.println(n);
                    }
                    else {
                        ind++;
                        pw.println(ind+1);
                    }
                }
            }
        }
        
        
        
        
        
        pw.flush();
        pw.close();
    }
}
