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 = 1e5 + 5;
char S[maxn], T[maxn];
int rSb[maxn], rTb[maxn], rSa[maxn], rTa[maxn];
int main() {
  scanf("%s", S + 1);
  int len = strlen(S + 1);
  for (int i = 1; i <= len; i++) {
    if (S[i] == 'C') S[i] = 'B';
    rSb[i] = rSb[i - 1];
    if (S[i] == 'B') {
      rSb[i]++;
      rSa[i] = 0;
    } else
      rSa[i] = rSa[i - 1] + 1;
  }
  scanf("%s", T + 1);
  len = strlen(T + 1);
  for (int i = 1; i <= len; i++) {
    if (T[i] == 'C') T[i] = 'B';
    rTb[i] = rTb[i - 1];
    if (T[i] == 'B') {
      rTb[i]++;
      rTa[i] = 0;
    } else
      rTa[i] = rTa[i - 1] + 1;
  }
  int Q, a, b, c, d;
  scanf("%d", &Q);
  while (Q--) {
    scanf("%d%d%d%d", &a, &b, &c, &d);
    int Sa = min(rSa[b], b - a + 1), Ta = min(rTa[d], d - c + 1);
    int Sb = rSb[b] - rSb[a - 1], Tb = rTb[d] - rTb[c - 1];
    if (Sb > Tb)
      printf("0");
    else if (Sb == Tb) {
      if (Sa >= Ta && (Sa - Ta) % 3 == 0)
        printf("1");
      else
        printf("0");
    } else if (Sb < Tb) {
      if (Sb > 0) {
        if (Sa == Ta) {
          if ((Tb - Sb) % 2 == 0)
            printf("1");
          else
            printf("0");
        } else if (Sa > Ta) {
          Sb += 2;
          if (Sb > Tb)
            printf("0");
          else if ((Tb - Sb) % 2 == 0)
            printf("1");
          else
            printf("0");
        } else
          printf("0");
      } else {
        if (Sa <= Ta)
          printf("0");
        else {
          Sb += 2;
          if (Sb > Tb)
            printf("0");
          else if ((Tb - Sb) % 2 == 0)
            printf("1");
          else
            printf("0");
        }
      }
    }
  }
  printf("\n");
  return 0;
}
