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 maxn = 2e5 + 7;
char S[maxn], T[maxn];
int Q, a, b, c, d;
int sumS[maxn], sumT[maxn];
int lastS[maxn], lastT[maxn];
int main() {
  scanf(" %s %s %d", S, T, &Q);
  for (int i = 0; S[i]; ++i) {
    sumS[i + 1] = sumS[i] + (S[i] == 'C' || S[i] == 'B');
    lastS[i + 1] = lastS[i];
    if (S[i] == 'B' || S[i] == 'C') lastS[i + 1] = i + 1;
  }
  for (int i = 0; T[i]; ++i) {
    sumT[i + 1] = sumT[i] + (T[i] == 'C' || T[i] == 'B');
    lastT[i + 1] = lastT[i];
    if (T[i] == 'B' || T[i] == 'C') lastT[i + 1] = i + 1;
  }
  while (Q--) {
    scanf("%d%d%d%d", &a, &b, &c, &d);
    int delta = sumT[d] - sumS[b] + sumS[a - 1] - sumT[c - 1];
    int delta2 = b - max(a - 1, lastS[b]) - (d - max(c - 1, lastT[d]));
    if (delta < 0 || delta % 2 != 0 || delta2 < 0 ||
        delta2 == 0 && !(sumS[b] - sumS[a - 1]) && delta > 0) {
      putchar('0');
      continue;
    }
    if (delta2 == 0 || delta > 0 || delta2 % 3 == 0)
      putchar('1');
    else
      putchar('0');
  }
  return 0;
}
