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 int MAXN = 100009;
char s1[MAXN], s2[MAXN], ans[MAXN];
int q, B1[MAXN], A1[MAXN], B2[MAXN], A2[MAXN];
int main() {
  memset(A1, 0, sizeof(A1));
  memset(B1, 0, sizeof(B1));
  memset(A2, 0, sizeof(A2));
  memset(B2, 0, sizeof(B2));
  scanf("%s %s", s1, s2);
  int len1 = strlen(s1);
  int len2 = strlen(s2);
  for (int i = 0; i < len1; ++i) {
    if (s1[i] == 'C') s1[i] = 'B';
    if (i != 0) B1[i + 1] = B1[i];
    if (s1[i] == 'B') ++B1[i + 1];
  }
  for (int i = 0; i < len2; ++i) {
    if (s2[i] == 'C') s2[i] = 'B';
    if (i != 0) B2[i + 1] = B2[i];
    if (s2[i] == 'B') ++B2[i + 1];
  }
  int tmp = 0;
  for (int i = 0; i < len1; ++i) {
    if (s1[i] == 'A')
      ++tmp;
    else
      tmp = 0;
    A1[i + 1] = tmp;
  }
  tmp = 0;
  for (int i = 0; i < len2; ++i) {
    if (s2[i] == 'A')
      ++tmp;
    else
      tmp = 0;
    A2[i + 1] = tmp;
  }
  scanf("%d", &q);
  for (int i = 0; i < q; ++i) {
    ans[i] = '0';
    int a, b, c, d;
    scanf("%d %d %d %d", &a, &b, &c, &d);
    int tmp1 = B1[b] - B1[a - 1];
    int tmp2 = B2[d] - B2[c - 1];
    int tmpp1 = min(b - a + 1, A1[b]);
    int tmpp2 = min(d - c + 1, A2[d]);
    if (tmp1 == tmp2) {
      if (tmpp1 >= tmpp2 && (tmpp1 - tmpp2) % 3 == 0) ans[i] = '1';
    } else if (tmp1 < tmp2 && (tmp2 - tmp1) % 2 == 0) {
      if (tmp1 == 0 && tmpp1 > tmpp2) ans[i] = '1';
      if (tmp1 != 0 && tmpp1 >= tmpp2) ans[i] = '1';
    }
  }
  puts(ans);
  return 0;
}
