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
# Author Name: Ajay Meena
# Codeforce : https://codeforces.com/profile/majay1638
import sys
import math
import bisect
import heapq
from bisect import bisect_right
from sys import stdin, stdout

# -------------- INPUT FUNCTIONS ------------------


def get_ints_in_variables(): return map(
    int, sys.stdin.readline().strip().split())


def get_int(): return int(sys.stdin.readline())


def get_ints_in_list(): return list(
    map(int, sys.stdin.readline().strip().split()))
def get_list_of_list(n): return [list(
    map(int, sys.stdin.readline().strip().split())) for _ in range(n)]


def get_string(): return sys.stdin.readline().strip()

# -------- SOME CUSTOMIZED FUNCTIONS-----------


def myceil(x, y): return (x + y - 1) // y

# -------------- SOLUTION FUNCTION ------------------


def Solution(s, a, b, p):
    # Write Your Code Here
    # 3 2 8
    # AABBBBAABB
    # [0,2,6,8]
    n = len(s)
    store = []
    prev = s[0]
    store.append(0)
    for i in range(1, n):
        if prev != s[i]:
            store.append(i)
            prev = s[i]
    store.append(n-1)
    money = 0
    ans = 0
    # print(store)
    m = len(store)-2
    if s[-1] != s[n-2]:
        m = len(store)-3
    for i in range(m, -1, -1):
        if s[store[i]] == "B":
            if b+money <= p:
                money += b
            else:
                ans = store[i+1]
                break
        else:
            if a+money <= p:
                money += a
            else:
                ans = store[i+1]
                break
    # print(store)
    print(ans+1)


def main():
    # Take input Here and Call solution function
    for _ in range(get_int()):
        a, b, p = get_ints_in_variables()
        s = get_string()
        Solution(s, a, b, p)


# calling main Function
if __name__ == '__main__':
    main()
