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 qa[100010] = {0}, qb[100010] = {0}, qla[100010] = {0};
int pa[100010] = {0}, pb[100010] = {0}, pla[100010] = {0};
char a[100010], b[100010];
int n, m;
int la, lb, l1, r1, l2, r2;
int main() {
  cin >> a >> b;
  cin >> n;
  la = strlen(a);
  lb = strlen(b);
  for (int i = 1; i <= la; i++) {
    if (a[i - 1] == 'A') {
      qa[i] = qa[i - 1] + 1;
      qla[i] = qla[i - 1];
      qb[i] = qb[i - 1];
    } else {
      qa[i] = qa[i - 1];
      qla[i] = i;
      qb[i] = qb[i - 1] + 1;
    }
  }
  for (int i = 1; i <= lb; i++) {
    if (b[i - 1] == 'A') {
      pa[i] = pa[i - 1] + 1;
      pla[i] = pla[i - 1];
      pb[i] = pb[i - 1];
    } else {
      pla[i] = i;
      pa[i] = pa[i - 1];
      pb[i] = pb[i - 1] + 1;
    }
  }
  for (int i = 0; i < n; i++) {
    cin >> l1 >> r1 >> l2 >> r2;
    int cna = qb[r1] - qb[l1 - 1], cnb = pb[r2] - pb[l2 - 1];
    int ans = 0;
    if (cnb == cna && cnb == 0 &&
        (r2 - l2 - r1 + l1 == 0 ||
         ((r2 - l2 - r1 + l1) % 3 == 0 && r2 - l2 - r1 + l1 < 0))) {
      ans = 1;
    }
    if (cna == 0 && cnb % 2 == 0 && cnb > 0 && (r2 - pla[r2] < r1 - l1 + 1)) {
      ans = 1;
    }
    if (cnb > cna && (cnb - cna) % 2 == 0 && cna > 0) {
      if ((r2 - pla[r2] <= r1 - qla[r1])) ans = 1;
    }
    if (cnb == cna && cnb > 0) {
      if (!((r1 - qla[r1] != r2 - pla[r2]) &&
            !((r1 - qla[r1] - r2 + pla[r2]) > 0 &&
              (r1 - qla[r1] - r2 + pla[r2]) % 3 == 0)))
        ans = 1;
    }
    cout << ans;
  }
  cout << endl;
  return 0;
}
