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 N = 1e5 + 50;
char str[N], ss[N];
int num1[N], num2[N], A1[N], A2[N];
int ans[N];
int main() {
  int i, j;
  scanf("%s%s", str + 1, ss + 1);
  int len1 = strlen(str + 1), len2 = strlen(ss + 1);
  num1[0] = num2[0] = A1[0] = A2[0] = 0;
  for (i = 1; i <= len1; i++) {
    if (str[i] != 'A')
      num1[i] = num1[i - 1] + 1, A1[i] = 0;
    else
      num1[i] = num1[i - 1], A1[i] = A1[i - 1] + 1;
  }
  for (i = 1; i <= len2; i++) {
    if (ss[i] != 'A')
      num2[i] = num2[i - 1] + 1, A2[i] = 0;
    else
      num2[i] = num2[i - 1], A2[i] = A2[i - 1] + 1;
  }
  int Q;
  scanf("%d", &Q);
  for (i = 0; i < Q; i++) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    int cnt1B = num1[b] - num1[a - 1], cnt1A = min(b - a + 1, A1[b]);
    int cnt2B = num2[d] - num2[c - 1], cnt2A = min(d - c + 1, A2[d]);
    if (cnt2B < cnt1B || cnt1A < cnt2A || ((cnt2B - cnt1B) % 2) ||
        cnt1B == 0 && cnt1A == cnt2A && cnt2B) {
      ans[i] = 0;
      continue;
    }
    cnt1A -= cnt2A;
    int needA = (cnt2B - cnt1B) / 2;
    ans[i] = 0;
    if (needA > 0 || needA == 0 && cnt1A % 3 == 0) ans[i] = 1;
  }
  for (i = 0; i < Q; i++) printf("%d", ans[i]);
  printf("\n");
  return 0;
}
