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 sb_count[100050];
int tb_count[100050];
int slast_b[100050];
int tlast_b[100050];
string s, t;
int calc(int &a, int &b, int &c, int &d) {
  int ps = max(a - 1, slast_b[b]);
  int pt = max(c - 1, tlast_b[d]);
  int len_s = b - ps;
  int len_t = d - pt;
  if (len_t > len_s) {
    return 0;
  }
  b -= len_t;
  d -= len_t;
  if (b < a) {
    return d >= c ? 0 : 1;
  }
  int scount = sb_count[b] - sb_count[a - 1];
  if (scount == 0 && d < c && (b - a + 1) % 3 == 0) {
    return 1;
  }
  int tcount = tb_count[d] - tb_count[c - 1];
  if ((b - ps) % 3) {
    scount += 2;
  }
  if (scount > tcount) {
    return 0;
  }
  if ((tcount - scount) % 2 == 1) {
    return 0;
  }
  return 1;
}
void solution() {
  int q;
  cin >> s >> t >> q;
  s = "*" + s;
  t = "*" + t;
  int last = 0;
  for (int i = 1; i < s.size(); i++) {
    sb_count[i] = sb_count[i - 1];
    if (s[i] == 'C' || s[i] == 'B') {
      sb_count[i]++;
      last = i;
    }
    slast_b[i] = last;
  }
  last = 0;
  for (int i = 1; i < t.size(); i++) {
    tb_count[i] = tb_count[i - 1];
    if (t[i] == 'C' || t[i] == 'B') {
      tb_count[i]++;
      last = i;
    }
    tlast_b[i] = last;
  }
  for (int i = 0; i < q; i++) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    printf("%d", calc(a, b, c, d));
  }
  cout << '\n';
}
int main() { solution(); }
