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
for _ in range(int(input())):
	a,b,p = list(map(int,input().split()))
	word = input()
	A,B = 0,0
	cost = []
	freq = []
	for i in range(len(word)):
		if word[i]=='A':
			if B!=0:
				freq.append(B)
				cost.append(b)
				B = 0
			A+=1
		else:
			if A!=0:
				freq.append(A)
				cost.append(a)
				A = 0
			B+=1
	if A!=0:
		freq.append(A)
		cost.append(a)
	if B!=0:
		freq.append(B)
		cost.append(b)
	flag = 0
	for i in cost:
		if p>=i:
			flag = 1
			break
	if flag==0:
		print(len(word))
	else:
		if word[-1]==word[-2]:
			cnt = 0
			freq = freq[-1::-1]
			cost = cost[-1::-1]
			for i in range(len(cost)):
				if cost[i]>p:
					break
				else:
					cnt+=freq[i]
					p-=cost[i]
			if len(word)==cnt:
				print(1)
			else:
				if cnt==0:
					print(len(word)-cnt)
				else:
					print(len(word)-cnt+1)
		else:
			cnt = 0
			freq = freq[-1::-1]
			cost = cost[-1::-1]
			for i in range(1,len(freq)):
				if cost[i]>p:
					break
				else:
					cnt+=freq[i]
					p-=cost[i]
			cnt+=1
			print(len(word)-cnt+1)