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>
char s[100001], t[100001];
int a[100001], b[100001], i, pa[100001], pb[100001], q, l, r, x, y, la[100001],
    lb[100001], za, zb;
int _(char c) { return c == 'A' ? 0 : 1; }
int mn(int x, int y) { return x < y ? x : y; }
int main() {
  scanf("%s\n%s\n", &s, &t);
  i = 0;
  while (s[i]) a[++a[0]] = _(s[i++]);
  i = 0;
  while (t[i]) b[++b[0]] = _(t[i++]);
  for (i = 1; i <= a[0]; i++) {
    pa[i] = pa[i - 1] + a[i];
    if (a[i] == 0)
      la[i] = la[i - 1] + 1;
    else
      la[i] = 0;
  }
  for (i = 1; i <= b[0]; i++) {
    pb[i] = pb[i - 1] + b[i];
    if (b[i] == 0)
      lb[i] = lb[i - 1] + 1;
    else
      lb[i] = 0;
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d%d%d", &l, &r, &x, &y);
    if (pa[r] - pa[l - 1] > pb[y] - pb[x - 1]) {
      printf("0");
      continue;
    };
    if ((pa[r] - pa[l - 1] - pb[y] + pb[x - 1]) % 2) {
      printf("0");
      continue;
    };
    za = mn(la[r], r - l + 1);
    zb = mn(lb[y], y - x + 1);
    if (pa[r] - pa[l - 1] == pb[y] - pb[x - 1]) {
      if (za == zb) {
        printf("1");
        continue;
      };
      if (za < zb) {
        printf("0");
        continue;
      };
      if ((za - zb) % 3) {
        printf("0");
        continue;
      };
      {
        printf("1");
        continue;
      };
    } else {
      if (pa[r] - pa[l - 1]) {
        if (za >= zb) {
          printf("1");
          continue;
        };
        {
          printf("0");
          continue;
        };
      } else {
        if (za > zb) {
          printf("1");
          continue;
        };
        {
          printf("0");
          continue;
        };
      }
    }
  }
}
