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 main() {
  std::string str1, str2;
  std::cin >> str1 >> str2;
  std::string ans = "";
  int as1[100010] = {0}, as2[100010] = {0}, bs1[100010] = {0},
      bs2[100010] = {0}, trailingAs1[100010] = {0}, trailingAs2[100010] = {0};
  int n = str1.length(), m = str2.length();
  for (int i = 1; i <= n; ++i) {
    as1[i] += as1[i - 1];
    bs1[i] += bs1[i - 1];
    trailingAs1[i] += trailingAs1[i - 1] + 1;
    if (str1[i - 1] == 'A')
      ++as1[i];
    else {
      ++bs1[i];
      trailingAs1[i] = 0;
    }
  }
  for (int i = 1; i <= m; ++i) {
    as2[i] += as2[i - 1];
    bs2[i] += bs2[i - 1];
    trailingAs2[i] += trailingAs2[i - 1] + 1;
    if (str2[i - 1] == 'A')
      ++as2[i];
    else {
      ++bs2[i];
      trailingAs2[i] = 0;
    }
  }
  int q;
  std::cin >> q;
  while (q--) {
    int l1, r1, l2, r2;
    std::cin >> l1 >> r1 >> l2 >> r2;
    int aas1 = as1[r1] - as1[l1 - 1];
    int aas2 = as2[r2] - as2[l2 - 1];
    int bbs1 = bs1[r1] - bs1[l1 - 1];
    int bbs2 = bs2[r2] - bs2[l2 - 1];
    int trailingA1 = std::min(trailingAs1[r1], r1 - l1 + 1);
    int trailingA2 = std::min(trailingAs2[r2], r2 - l2 + 1);
    if (trailingA1 < trailingA2 || bbs1 > bbs2) {
      ans += "0";
    } else if (bbs1 == bbs2) {
      if (trailingA1 >= trailingA2) {
        if (trailingA1 % 3 == trailingA2 % 3) {
          ans += "1";
        } else
          ans += "0";
      } else {
        ans += "0";
      }
    } else if (bbs1 % 2 == bbs2 % 2) {
      if (bbs1 == 0) {
        --trailingA1;
        if (trailingA1 < trailingA2) {
          ans += "0";
        } else {
          ans += "1";
        }
      } else {
        ans += "1";
      }
    } else {
      ans += "0";
    }
  }
  std::cout << ans;
}
