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>
const int MAXN = 100100;
char S[MAXN];
char T[MAXN];
int aS[MAXN], aT[MAXN];
int bS[MAXN], bT[MAXN];
bool go() {
  int leftS, rightS, leftT, rightT;
  scanf("%d %d %d %d", &leftS, &rightS, &leftT, &rightT);
  int lengthS = rightS - leftS + 1;
  int lengthT = rightT - leftT + 1;
  int as = aS[rightS];
  if (as > lengthS) as = lengthS;
  int at = aT[rightT];
  if (at > lengthT) at = lengthT;
  int bs = bS[rightS] - bS[leftS - 1];
  int bt = bT[rightT] - bT[leftT - 1];
  if (bs % 2 != bt % 2) {
    return false;
  }
  if (bs > bt) {
    return false;
  }
  if (as < at) {
    return false;
  }
  if (bs == bt) {
    if (as % 3 != at % 3) {
      return false;
    }
  } else {
    if (bs == 0 && as == at) {
      return false;
    }
  }
  return true;
}
void goA(char *s, int *ar) {
  int run = 0;
  for (int i = 1; s[i]; ++i) {
    if (s[i] == 'A')
      ++run;
    else
      run = 0;
    ar[i] = run;
  }
}
void goB(char *s, int *ar) {
  int run = 0;
  for (int i = 1; s[i]; ++i) {
    if (s[i] == 'B' || s[i] == 'C') ++run;
    ar[i] = run;
  }
}
int main() {
  scanf("%s", S + 1);
  scanf("%s", T + 1);
  goA(S, aS);
  goA(T, aT);
  goB(S, bS);
  goB(T, bT);
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; ++i) {
    if (go()) {
      printf("1");
    } else {
      printf("0");
    }
  }
  printf("\n");
}
