923_D. Picking Strings
Alice has a string consisting of characters 'A', 'B' and 'C'. Bob can use the following transitions on any substring of our string in any order any number of times: 

  * A <image> BC
  * B <image> AC
  * C <image> AB
  * AAA <image> empty string 



Note that a substring is one or more consecutive characters. For given queries, determine whether it is possible to obtain the target string from source.

Input

The first line contains a string S (1 ≤ |S| ≤ 105). The second line contains a string T (1 ≤ |T| ≤ 105), each of these strings consists only of uppercase English letters 'A', 'B' and 'C'.

The third line contains the number of queries Q (1 ≤ Q ≤ 105).

The following Q lines describe queries. The i-th of these lines contains four space separated integers ai, bi, ci, di. These represent the i-th query: is it possible to create T[ci..di] from S[ai..bi] by applying the above transitions finite amount of times?

Here, U[x..y] is a substring of U that begins at index x (indexed from 1) and ends at index y. In particular, U[1..|U|] is the whole string U.

It is guaranteed that 1 ≤ a ≤ b ≤ |S| and 1 ≤ c ≤ d ≤ |T|.

Output

Print a string of Q characters, where the i-th character is '1' if the answer to the i-th query is positive, and '0' otherwise.

Example

Input

AABCCBAAB
ABCB
5
1 3 1 2
2 2 2 4
7 9 1 1
3 4 2 3
4 5 1 3


Output

10011

Note

In the first query we can achieve the result, for instance, by using transitions <image>.

The third query asks for changing AAB to A — but in this case we are not able to get rid of the character 'B'.
```python
def pref_counts(string, char):
    result = [0]
    for c in string:
        result.append(result[-1] + (c == char))
    return result

def left_counts(string, char):
    result = [0]
    for c in string:
        result.append(result[-1] + 1 if c == char else 0)
    return result

s = input().replace("C", "B")
t = input().replace("C", "B")
s_bcounts, t_bcounts = pref_counts(s, "B"), pref_counts(t, "B")
s_lcounts, t_lcounts = left_counts(s, "A"), left_counts(t, "A")
for i in range(int(input())):
    a, b, c, d = list(map(int, input().split()))
    s_b = s_bcounts[b] - s_bcounts[a - 1]
    t_b = t_bcounts[d] - t_bcounts[c - 1]
    s_a = min(s_lcounts[b], b - a + 1)
    t_a = min(t_lcounts[d], d - c + 1)
    answer = False
    if s_b & 1 == t_b & 1:
        if s_b < t_b:
            answer = s_a - (s_b == 0) >= t_a
        elif s_b == t_b:
            answer = s_a >= t_a and (s_a - t_a) % 3 == 0
    print(int(answer), end = "")


```