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;
const double g = 10.0, eps = 1e-12;
const int N = 100000 + 10, maxn = 2000 + 10, inf = 0x3f3f3f3f,
          INF = 0x3f3f3f3f3f3f3f3f;
char s[N], p[N];
int sums[N], sump[N];
int lens, lenp;
int cal(int l, int r, int op) {
  int tel = l - 1, ter = r + 1;
  while (tel < ter - 1) {
    int m = (tel + ter) >> 1;
    if (op == 1) {
      if (sums[m] - sums[r + 1] >= r - m + 1)
        ter = m;
      else
        tel = m;
    } else {
      if (sump[m] - sump[r + 1] >= r - m + 1)
        ter = m;
      else
        tel = m;
    }
  }
  return tel;
}
bool solve(int l1, int r1, int l2, int r2) {
  int lastb1 = cal(l1, r1, 1);
  int lastb2 = cal(l2, r2, 2);
  int a1 = r1 - lastb1, a2 = r2 - lastb2;
  int b1 = lastb1 - l1 + 1, b2 = lastb2 - l2 + 1;
  b1 -= (sums[l1] - sums[lastb1 + 1]);
  b2 -= (sump[l2] - sump[lastb2 + 1]);
  if (a1 > a2) {
    int te1 = b1 + 2, te2 = b2;
    int te3 = b1, te4 = b2;
    if (te1 <= te2 && (te2 - te1) % 2 == 0)
      return 1;
    else if ((a1 - a2) % 3 == 0 && te3 <= te4 && (te4 - te3) % 2 == 0)
      return 1;
    else
      return 0;
  } else if (a1 == a2) {
    if (b1 == 0 && b2 != 0)
      return 0;
    else if (b1 <= b2 && (b2 - b1) % 2 == 0)
      return 1;
  }
  return 0;
}
int main() {
  scanf("%s%s", s + 1, p + 1);
  lens = strlen(s + 1), lenp = strlen(p + 1);
  for (int i = lens; i >= 0; i--) {
    if (s[i] == 'A')
      sums[i] = sums[i + 1] + 1;
    else
      sums[i] = sums[i + 1];
  }
  for (int i = lenp; i >= 0; i--) {
    if (p[i] == 'A')
      sump[i] = sump[i + 1] + 1;
    else
      sump[i] = sump[i + 1];
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int l1, r1, l2, r2;
    scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
    if (solve(l1, r1, l2, r2))
      printf("1");
    else
      printf("0");
  }
  puts("");
  return 0;
}
