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 PI = acos(-1.0);
const int INF = 1000 * 1000 * 1000 + 7;
const long long LINF = INF * (long long)INF;
const int MAX = 100100;
struct Fen {
  int A[MAX];
  int n;
  void init(int n) { this->n = n; }
  void add(int x, int val) {
    for (; x < n; x = x | (x + 1)) {
      A[x] += val;
    }
  }
  int get(int x) {
    int res = 0;
    for (; x >= 0; x = (x & (x + 1)) - 1) {
      res += A[x];
    }
    return res;
  }
  int get(int l, int r) {
    int res = get(r);
    res -= get(l - 1);
    return res;
  }
} F1, F2;
char S[MAX];
char T[MAX];
int C1[MAX];
int C2[MAX];
bool solve(int a, int b, int c, int d) {
  int s1 = F1.get(a, b);
  int s2 = F2.get(c, d);
  if (s1 % 2 != s2 % 2) return false;
  if (s1 > s2) return false;
  int cnt1 = C1[b];
  int cnt2 = C2[d];
  cnt1 = min(cnt1, b - a + 1);
  cnt2 = min(cnt2, d - c + 1);
  if (s1 == s2) {
    if (cnt1 < cnt2) return false;
    if (cnt1 % 3 == cnt2 % 3) return true;
    return false;
  }
  if (s1 == 0) {
    if (cnt1 == cnt2) return false;
  }
  if (cnt1 < cnt2) return false;
  return true;
}
int main() {
  scanf("%s", S);
  scanf("%s", T);
  int n = strlen(S);
  int m = strlen(T);
  F1.init(n);
  F2.init(m);
  for (int i = (0); i < (n); i++) {
    if (S[i] == 'C' || S[i] == 'B') F1.add(i, 1);
  }
  for (int i = (0); i < (m); i++) {
    if (T[i] == 'C' || T[i] == 'B') F2.add(i, 1);
  }
  if (S[0] == 'A')
    C1[0] = 1;
  else
    C1[0] = 0;
  for (int i = (1); i < (n); i++) {
    C1[i] = C1[i - 1];
    if (S[i] == 'A')
      C1[i]++;
    else
      C1[i] = 0;
  }
  if (T[0] == 'A')
    C2[0] = 1;
  else
    C2[0] = 0;
  for (int i = (1); i < (m); i++) {
    C2[i] = C2[i - 1];
    if (T[i] == 'A')
      C2[i]++;
    else
      C2[i] = 0;
  }
  int q;
  scanf("%d", &q);
  for (int i = (0); i < (q); i++) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    a--;
    b--;
    c--;
    d--;
    if (solve(a, b, c, d))
      printf("1");
    else
      printf("0");
  }
  printf("\n");
}
