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 sys,math,bisect
from random import randint
inf = float('inf')
mod = (10**9)+7
"========================================"
def lcm(a,b):
    return int((a/math.gcd(a,b))*b)
def gcd(a,b):
    return int(math.gcd(a,b))
def tobinary(n):
    return bin(n)[2:]
def binarySearch(a,x):
    i = bisect.bisect_left(a,x)
    if i!=len(a) and a[i]==x:
        return i
    else:
        return -1
def lowerBound(a, x):
    i = bisect.bisect_left(a, x)
    if i:
        return (i-1)
    else:
        return -1
def upperBound(a,x):
    i = bisect.bisect_right(a,x)
    if i!= len(a)+1 and a[i-1]==x:
        return (i-1)
    else:
        return -1
def primesInRange(n):
    ans = []
    prime = [True for i in range(n+1)]
    p = 2
    while (p * p <= n):
        if (prime[p] == True):
            for i in range(p * p, n+1, p):
                prime[i] = False
        p += 1
    for p in range(2, n+1):
        if prime[p]:
            ans.append(p)
    return ans
def primeFactors(n):
    factors = []
    while n % 2 == 0:
        factors.append(2)
        n = n // 2
    for i in range(3,int(math.sqrt(n))+1,2):
        while n % i== 0:
            factors.append(i)
            n = n // i
    if n > 2:
        factors.append(n)
    return factors
def isPrime(n,k=5):
    if (n <2):
        return True
    for i in range(0,k):
        a = randint(1,n-1)
        if(pow(a,n-1,n)!=1):
            return False
    return True
"========================================="
"""
n = int(input())
n,k = map(int,input().split())
arr = list(map(int,input().split()))
"""
from collections import deque,defaultdict,Counter
import heapq,string

for _ in range(int(input())):
    b,a,p=map(int,input().split())
    s=input()
    n=len(s)
    if len(s)==1:
        print(1)
        continue
    l=0
    ans=len(s)
    for i in range(n-2,-1,-1):
        if i==0 or s[i]!=s[i-1]:
            if s[i]=='A':
                l+=b
            else:
                l+=a
            if l<=p:
                ans = i+1
    print(ans)
