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 INF = 0x3f3f3f3f;
const long long INFF = 0x3f3f3f3f3f3f3f3fll;
const long long M = 1e9 + 7;
const long long maxn = 3e5 + 7;
const double eps = 0.00000001;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
template <typename T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
template <typename T>
inline T powMM(T a, T b) {
  T ret = 1;
  for (; b; b >>= 1ll, a = 1ll * a * a % M)
    if (b & 1) ret = 1ll * ret * a % M;
  return ret;
}
int n, m;
int i, j, k;
char S[maxn], T[maxn];
int TA[maxn], TB[maxn], SA[maxn], SB[maxn];
char ans[maxn];
int check(int CA, int CB, int DA, int DB) {
  if (CA < DA) return 0;
  if (CA == DA && !CB && DB) return 0;
  if ((CA - DA) % 3) CB += 2;
  if (DB < CB) return 0;
  if ((DB - CB) & 1) return 0;
  return 1;
}
int solve(int CA, int CB, int DA, int DB) {
  int ret = check(CA, CB, DA, DB);
  printf("  %d %d %d %d; %d\n", CA, CB, DA, DB, ret);
  return ret;
}
int main() {
  scanf("%s%s", S + 1, T + 1);
  n = strlen(S + 1);
  for (i = 1; i <= n; i++) {
    if (S[i] == 'A')
      SA[i] = SA[i - 1] + 1, SB[i] = SB[i - 1];
    else
      SB[i] = SB[i - 1] + 1;
  }
  n = strlen(T + 1);
  for (i = 1; i <= n; i++) {
    if (T[i] == 'A')
      TA[i] = TA[i - 1] + 1, TB[i] = TB[i - 1];
    else
      TB[i] = TB[i - 1] + 1;
  }
  int C;
  scanf("%d", &C);
  for (i = 0; i < C; i++) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    int CA = min(SA[b], b - a + 1), CB = SB[b] - SB[a - 1];
    int DA = min(TA[d], d - c + 1), DB = TB[d] - TB[c - 1];
    ans[i] = check(CA, CB, DA, DB) + '0';
  }
  ans[C] = 0;
  puts(ans);
}
