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 $s_i = \texttt{A}$, if there is a bus station at $i$-th crossroad, and $s_i = \texttt{B}$, 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 $i, i+1, \ldots, j-1$ 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 $s_t = \texttt{A}$ for all $i \le t < j$. 

If for two crossroads $i$ and $j$ for all crossroads $i, i+1, \ldots, j-1$ 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 $s_t = \texttt{B}$ for all $i \le t < j$.

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 $t$ ($1 \le t \le 10^4$).

The first line of each test case consists of three integers $a, b, p$ ($1 \le a, b, p \le 10^5$) — 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 $s_i = \texttt{A}$, if there is a bus station at $i$-th crossroad, and $s_i = \texttt{B}$, if there is a tram station at $i$-th crossroad ($2 \le |s| \le 10^5$).

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

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 'import sys\nimport math\nfrom collections import defaultdict\nfrom collections import deque\nfrom itertools import combinations\nfrom itertools import permutations\ninput = lambda : sys.stdin.readline().rstrip()\nread = lambda : list(map(int, input().split()))\ngo = lambda : 1/0\ndef write(*args, sep="\\n"):\n  for i in args:\n    sys.stdout.write("{}{}".format(i, sep))\nINF = float(\'inf\')\nMOD = int(1e9 + 7)\nYES = "YES"\nNO = "NO"\n\nfor _ in range(int(input())):\n  try:\n    a, b, p = read()\n    s = input()\n\n    stack = [[s[0], 1]]\n\n    for i in s[1:-1]:\n      if i == stack[-1][0]:\n        stack[-1][1] += 1\n        \n      else:\n        stack.append([i, 1])\n    \n\n\n    ans = len(s) \n    temp = []\n    # print(stack)\n\n    if p < a and p < b:\n      print(len(s))\n      go() \n    \n    while stack:\n      i, j = stack[-1]\n      stack.pop()\n      if i == \'A\' and p >= a:\n        p -= a\n        ans -= j\n        temp.append(j)\n      elif i == \'A\' and p < a:\n        break\n      if i == \'B\' and p >= b:\n        p -= b \n        ans -= j \n        temp.append(j)\n      elif i == \'B\' and p < b:\n        break\n    \n    print(ans)\n      \n\n  except ZeroDivisionError:\n    continue\n\n  except Exception as e:\n    print(e)\n    continue'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 15369