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'.
#include <bits/stdc++.h>
using namespace std;
const long double EPS = 1e-9, PI = acos(-1.);
const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const int N = 1e5 + 5, SQ = 450;
char s[N], t[N];
int q, a, b, c, d;
int ans[N];
int bs[N], bt[N], as[N], at[N];
int main() {
  scanf("%s%s", s + 1, t + 1);
  for (int i = 1; s[i]; i++) bs[i] = bs[i - 1] + (s[i] != 'A');
  for (int i = 1; t[i]; i++) bt[i] = bt[i - 1] + (t[i] != 'A');
  for (int i = 1; s[i]; i++) as[i] = (s[i] != 'A') ? 0 : as[i - 1] + 1;
  for (int i = 1; t[i]; i++) at[i] = (t[i] != 'A') ? 0 : at[i - 1] + 1;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    scanf("%d%d%d%d", &a, &b, &c, &d);
    int vs = bs[b] - bs[a - 1];
    int ts = bt[d] - bt[c - 1];
    int ab = min(b - a + 1, as[b]);
    int ad = min(d - c + 1, at[d]);
    int ok = 1;
    if (vs > ts) ok = 0;
    if (vs % 2 != ts % 2) ok = 0;
    if (ab < ad) ok = 0;
    if (ab > ad and (ab - ad) % 3 and vs + 2 > ts) ok = 0;
    if (ab == ad and ab == b - a + 1 and ad != d - c + 1) ok = 0;
    printf("%d", ok);
  }
  printf("\n");
  return 0;
}
