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
from os import path
import sys
# mod = int(1e9 + 7)
# import re  # can use multiple splits
from math import ceil, floor,gcd,log2
from collections import defaultdict , Counter
# from bisect import bisect_left, bisect_right
#popping from the end is less taxing,since you don't have to shift any elements
maxx = float('inf')

if (path.exists('input.txt')): 
	#------------------Sublime--------------------------------------#
	sys.stdin=open('input.txt','r');sys.stdout=open('output.txt','w');
	I = lambda :int(sys.stdin.buffer.readline())
	tup= lambda : map(int , sys.stdin.buffer.readline().split())
	lint = lambda :[int(x) for x in sys.stdin.buffer.readline().split()]
	S = lambda: sys.stdin.readline().replace('\n', '').strip()
	# def grid(r, c): return [lint() for i in range(r)]
	# def debug(*args, c=6): print('\033[3{}m'.format(c), *args, '\033[0m', file=sys.stderr)
	stpr = lambda x : sys.stdout.write(f'{x}' + '\n')
	star = lambda x: print(' '.join(map(str, x)))


else:
	#------------------PYPY FAst I/o--------------------------------#
	I = lambda :int(sys.stdin.buffer.readline())
	tup= lambda : map(int , sys.stdin.buffer.readline().split())
	lint = lambda :[int(x) for x in sys.stdin.buffer.readline().split()]
	S = lambda: sys.stdin.readline().replace('\n', '').strip()
	# def grid(r, c): return [lint() for i in range(r)]
	stpr = lambda x : sys.stdout.write(f'{x}' + '\n')
	star = lambda x: print(' '.join(map(str, x)))
input = sys.stdin.readline
for _ in range(int(input())):
	a ,b ,p= map(int , input().split())
	s = list(input().strip())
	n , t = len(s) , 0
	if len(s) > 1 :
		if s[-1] != s[-2] :
			t = 1
			s.pop()
			n-=1
	st,exp =[] , 0
	for i in range(n):
		if not st:
			st.append(s[i])
			if s[i] == 'A':
				exp+=a
			else:exp+=b
		else:
			if s[i] != st[-1]:
				if s[i] =='A':
					exp+=a
				else:exp+=b
				st.append(s[i])
	# print(exp)
	if exp <= p:
		print(1)
	else:
		i , ss,F = 0 , [],True
		while  i < n:
			if len(ss) == 0:
				ss.append(s[i])
				if s[i] =='A':
					exp-=a
				else:exp-=b
				ss.append(s[i])
			else:
				if ss[-1] != s[i]:
					if exp <= p :
						print(i+1)
						F = False
						break
					if s[i] =='A':
						exp-=a
					else:exp-=b
					ss.append(s[i])
			i+=1
		if F:
			if t :
				print(i+1)
			else:print(i)