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;
char s[100001], t[100001];
int qza[100001], n, qzb[100001], lefta[100001], leftb[100001], left1a[100001],
    left1b[100001];
int main() {
  scanf("%s\n", &s);
  int len1 = strlen(s);
  scanf("%s\n", &t);
  int len2 = strlen(t);
  for (int i = 0; i <= len1 - 1; i++) {
    qza[i + 1] = qza[i];
    if (s[i] > 'A') {
      qza[i + 1] = qza[i] + 1;
    }
  }
  for (int i = 0; i <= len2 - 1; i++) {
    qzb[i + 1] = qzb[i];
    if (t[i] > 'A') {
      qzb[i + 1] = qzb[i] + 1;
    }
  }
  for (int i = 0; i <= len1 - 1; i++) {
    if (s[i] == 'A') {
      lefta[i + 1] = lefta[i];
    } else {
      lefta[i + 1] = i + 1;
    }
    if (s[i] > 'A') {
      left1a[i + 1] = left1a[i];
    } else {
      left1a[i + 1] = i + 1;
    }
  }
  for (int i = 0; i <= len2 - 1; i++) {
    if (i == 16) {
      len2++;
      len2--;
    }
    if (t[i] == 'A') {
      leftb[i + 1] = leftb[i];
    } else {
      leftb[i + 1] = i + 1;
    }
    if (t[i] > 'A') {
      left1b[i + 1] = left1b[i];
    } else {
      left1b[i + 1] = i + 1;
    }
  }
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    if (i == 11) {
      n++;
      n--;
    }
    int a, b, a1, b1;
    scanf("%d%d%d%d", &a, &b, &a1, &b1);
    int numbera = qza[b] - qza[a - 1];
    int numberb = qzb[b1] - qzb[a1 - 1];
    if ((numbera % 2 != numberb % 2) || (numbera > numberb)) {
      printf("0");
      continue;
    }
    int numa = b - lefta[b];
    int numb = b1 - leftb[b1];
    if (lefta[b] < a) numa = b - a + 1;
    if (leftb[b1] < a1) numb = b1 - a1 + 1;
    if (numa < numb) {
      printf("0");
      continue;
    }
    if (numbera == numberb && numa % 3 != numb % 3) {
      printf("0");
      continue;
    }
    if (numa == numb && numbera == 0 && numberb > 0) {
      printf("0");
      continue;
    }
    printf("1");
  }
}
