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>
int s1[100001], s2[100001], a, b, c, d, ansa[100001], q, n, m;
char S[100001], T[100001];
int main() {
  int i, j, SA, SB, TA, TB, l, r, mid, q1;
  scanf("%s\n%s\n", S, T);
  n = strlen(S);
  m = strlen(T);
  for (i = 1; i <= n; i++) {
    if (S[i - 1] != 'A')
      s1[i] = s1[i - 1] + 1;
    else
      s1[i] = s1[i - 1];
  }
  for (i = 1; i <= m; i++) {
    if (T[i - 1] != 'A')
      s2[i] = s2[i - 1] + 1;
    else
      s2[i] = s2[i - 1];
  }
  scanf("%d", &q);
  q1 = q;
  i = 1;
  while (q1 >= 1) {
    scanf("%d %d %d %d", &a, &b, &c, &d);
    SB = s1[b] - s1[a - 1];
    TB = s2[d] - s2[c - 1];
    l = a;
    r = b;
    j = b + 1;
    while (l <= r) {
      mid = (l + r) / 2;
      if (s1[b] - s1[mid - 1] == 0) {
        j = mid;
        r = mid - 1;
      } else {
        l = mid + 1;
      }
    }
    SA = b - j + 1;
    l = c;
    r = d;
    j = d + 1;
    while (l <= r) {
      mid = (l + r) / 2;
      if (s2[d] - s2[mid - 1] == 0) {
        j = mid;
        r = mid - 1;
      } else {
        l = mid + 1;
      }
    }
    TA = d - j + 1;
    if (SB <= TB && (TB - SB) % 2 == 0) {
      if (SB == TB) {
        if (SA >= TA && (SA - TA) % 3 == 0) {
          ansa[i] = 1;
        }
      } else {
        if (SB == 0) {
          if (SA > TA) {
            ansa[i] = 1;
          }
        } else {
          if (SA >= TA) {
            ansa[i] = 1;
          }
        }
      }
    }
    q1--;
    i++;
  }
  for (i = 1; i <= q; i++) printf("%d", ansa[i]);
}
