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 int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};
const double EPS = 1e-9;
const int MOD = 1000 * 1000 * 1000 + 7;
const int INF = 2000 * 1000 * 1000;
const int MAXN = 100010;
template <typename T>
inline T sqr(T n) {
  return n * n;
}
char s[MAXN], t[MAXN];
int sumS[MAXN], sumT[MAXN];
int cntAS[MAXN], cntAT[MAXN];
int main() {
  scanf("%s\n", s + 1);
  scanf("%s\n", t + 1);
  for (int i = 1; s[i]; i++) {
    sumS[i] = sumS[i - 1] + (s[i] == 'B' || s[i] == 'C');
  }
  for (int i = 1; t[i]; i++) {
    sumT[i] = sumT[i - 1] + (t[i] == 'B' || t[i] == 'C');
  }
  for (int i = 1; s[i]; i++) {
    if (s[i] == 'A') {
      cntAS[i] = cntAS[i - 1] + 1;
    } else {
      cntAS[i] = 0;
    }
  }
  for (int i = 1; t[i]; i++) {
    if (t[i] == 'A') {
      cntAT[i] = cntAT[i - 1] + 1;
    } else {
      cntAT[i] = 0;
    }
  }
  int n;
  scanf("%d", &n);
  while (n-- > 0) {
    int ls, rs, lt, rt;
    scanf("%d%d%d%d", &ls, &rs, &lt, &rt);
    int sBC = sumS[rs] - sumS[ls - 1];
    int tBC = sumT[rt] - sumT[lt - 1];
    int sA = min(cntAS[rs], rs - ls + 1);
    int tA = min(cntAT[rt], rt - lt + 1);
    if (sA < tA) {
      putchar('0');
      continue;
    }
    if (sBC % 2 != tBC % 2) {
      putchar('0');
      continue;
    }
    if (sBC == tBC) {
      if (sA % 3 == tA % 3) {
        putchar('1');
      } else {
        putchar('0');
      }
      continue;
    }
    if (sBC != 0 && sBC < tBC) {
      if (sA >= tA) {
        putchar('1');
      } else {
        putchar('0');
      }
      continue;
    }
    if (sBC == 0 && tBC != 0) {
      if (sA > tA) {
        putchar('1');
      } else {
        putchar('0');
      }
      continue;
    }
    putchar('0');
  }
  putchar('\n');
  return 0;
}
