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;
int main() {
  cin.tie(NULL);
  iostream::sync_with_stdio(false);
  string s;
  cin >> s;
  string t;
  cin >> t;
  vector<int> stars, start;
  vector<int> as, at;
  stars.push_back(0);
  as.push_back(0);
  for (int i = 0; i < s.size(); ++i) {
    if (s[i] == 'B' || s[i] == 'C') {
      stars.push_back(stars.back() + 1);
      as.push_back(0);
    } else {
      stars.push_back(stars.back());
      as.push_back(as.back() + 1);
    }
  }
  start.push_back(0);
  at.push_back(0);
  for (int i = 0; i < t.size(); ++i) {
    if (t[i] == 'B' || t[i] == 'C') {
      start.push_back(start.back() + 1);
      at.push_back(0);
    } else {
      start.push_back(start.back());
      at.push_back(at.back() + 1);
    }
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    int ls, rs, lt, rt;
    cin >> ls >> rs >> lt >> rt;
    int star_cnt_s = stars[rs] - stars[ls - 1];
    int star_cnt_t = start[rt] - start[lt - 1];
    int a_cnt_s = min(as[rs], rs - ls + 1);
    int a_cnt_t = min(at[rt], rt - lt + 1);
    if (star_cnt_t < star_cnt_s) {
      cout << '0';
      continue;
    }
    if (star_cnt_t % 2 != star_cnt_s % 2) {
      cout << '0';
      continue;
    }
    if (a_cnt_s < a_cnt_t) {
      cout << '0';
      continue;
    }
    if (star_cnt_s == star_cnt_t) {
      if (a_cnt_s % 3 == a_cnt_t % 3) {
        cout << '1';
      } else {
        cout << '0';
      }
    } else {
      if (star_cnt_s == 0) {
        if (a_cnt_s - 1 < a_cnt_t) {
          cout << '0';
        } else {
          cout << '1';
        }
      } else {
        cout << '1';
      }
    }
  }
  return 0;
}
