Let us consider the following operations on a string consisting of A and B:
 - Select a character in a string. If it is A, replace it with BB. If it is B, replace with AA.
 - Select a substring that is equal to either AAA or BBB, and delete it from the string.
For example, if the first operation is performed on ABA and the first character is selected, the string becomes BBBA.
If the second operation is performed on BBBAAAA and the fourth through sixth characters are selected, the string becomes BBBA.
These operations can be performed any number of times, in any order.
You are given two string S and T, and q queries a_i, b_i, c_i, d_i.
For each query, determine whether S_{a_i} S_{{a_i}+1} ... S_{b_i}, a substring of S, can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, a substring of T.

-----Constraints-----
 - 1 \leq |S|, |T| \leq 10^5
 - S and T consist of letters A and B.
 - 1 \leq q \leq 10^5
 - 1 \leq a_i \leq b_i \leq |S|
 - 1 \leq c_i \leq d_i \leq |T|

-----Input-----
Input is given from Standard Input in the following format:
S
T
q
a_1 b_1 c_1 d_1
...
a_q b_q c_q d_q

-----Output-----
Print q lines. The i-th line should contain the response to the i-th query. If S_{a_i} S_{{a_i}+1} ... S_{b_i} can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, print YES. Otherwise, print NO.

-----Sample Input-----
BBBAAAABA
BBBBA
4
7 9 2 5
7 9 1 4
1 7 2 5
1 7 2 4

-----Sample Output-----
YES
NO
YES
NO

The first query asks whether the string ABA can be made into BBBA.
As explained in the problem statement, it can be done by the first operation.
The second query asks whether ABA can be made into BBBB, and the fourth query asks whether BBBAAAA can be made into BBB.
Neither is possible.
The third query asks whether the string BBBAAAA can be made into BBBA.
As explained in the problem statement, it can be done by the second operation.

```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 = "# 【考察】\n# (1) A と B は入れ替え可能：\n#   AB -> AAA -> BBAA -> BBBBA -> BA\n# (2) AB と BA は空文字に置換可能：\n#   AB -> AAA -> ''\n# (3) 一つの文字から3つずつ同じ文字を増やしていくことが可能\n#   A -> BB -> AAB -> AAAA\n# 以上の点より、A・Bの個数の差のmod3の値が本質的だとわかる。\n\n\ndef cumsum_a(st):\n    cs = [0] * len(st)\n    for i, s in enumerate(st):\n        if s == 'A':\n            cs[i] = 1\n        if i > 0:\n            cs[i] += cs[i - 1]\n    return cs\n\n\ndef main():\n    S = input()\n    T = input()\n    Q = int(input())\n    query_list = [list(map(lambda x: int(x) - 1, input().split())) for _ in range(Q)]\n    # Aの個数の累積和をとる（bはaの個数からわかる）\n    s_cs_a, t_cs_a = cumsum_a(S), cumsum_a(T)\n    for query in query_list:\n        s_left, s_right, t_left, t_right = query\n        s_count_a = s_cs_a[s_right]\n        if s_left > 0:\n            s_count_a -= s_cs_a[s_left - 1]\n        s_count_b = s_right - s_left + 1 - s_count_a\n        t_count_a = t_cs_a[t_right]\n        if t_left > 0:\n            t_count_a -= t_cs_a[t_left - 1]\n        t_count_b = t_right - t_left + 1 - t_count_a\n        # 個数の差のmod3の値で比較する\n        s_diff = (s_count_a - s_count_b) % 3\n        t_diff = (t_count_a - t_count_b) % 3\n        if s_diff == t_diff:\n            print('YES')\n        else:\n            print('NO')\n\n\ndef __starting_point():\n    main()\n__starting_point()"
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 14195