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>
const int N = 1e5 + 7;
char s[N], t[N];
int n, m, sa[N], sb[N], ta[N], tb[N];
inline int read() {
  int x = 0;
  char c = getchar();
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  return x;
}
int main() {
  std::cin >> (s + 1) >> (t + 1);
  n = strlen(s + 1), m = strlen(t + 1);
  for (int i = 1; i <= n; ++i) {
    sb[i] = sb[i - 1];
    if (s[i] == 'A')
      sa[i] = sa[i - 1] + 1;
    else
      ++sb[i];
  }
  for (int i = 1; i <= m; ++i) {
    tb[i] = tb[i - 1];
    if (t[i] == 'A')
      ta[i] = ta[i - 1] + 1;
    else
      ++tb[i];
  }
  int q;
  std::cin >> q;
  while (q-- > 0) {
    int a = read(), b = read(), c = read(), d = read(), ans = 1;
    int db = tb[d] - tb[c - 1] - sb[b] + sb[a - 1];
    int Sa = std::min(b - a + 1, sa[b]), Ta = std::min(d - c + 1, ta[d]);
    if (db < 0 || Sa < Ta || db % 2 == 1) ans = 0;
    bool pd = (Sa - Ta) % 3 > 0;
    if (db - pd * 2 < 0) ans = 0;
    if (sb[b] - sb[a - 1] == 0 && db && Sa == Ta) ans = 0;
    printf("%d", ans);
  }
  return 0;
}
