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 + 10;
const int maxm = 1e5 + 100;
const double eps = 1e-6;
const long long MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
char s1[maxn], s2[maxn];
int sux1[maxn], sux2[maxn], pre1[maxn], pre2[maxn], ans[maxn];
bool Judge(int l1, int r1, int l2, int r2) {
  int a = pre1[r1] - pre1[l1 - 1], c = pre2[r2] - pre2[l2 - 1];
  int b = min(r1 - l1 + 1, sux1[r1]), d = min(r2 - l2 + 1, sux2[r2]);
  if (a > c || abs(a - c) % 2) return false;
  if (a == c) {
    if (b < d) return false;
    return (b - d) % 3 == 0;
  }
  if (a == 0) return b > d;
  return b >= d;
}
int main() {
  scanf("%s%s", s1 + 1, s2 + 1);
  int len1 = strlen(s1 + 1), len2 = strlen(s2 + 1);
  for (int i = 1; i <= len1; i++) {
    pre1[i] = pre1[i - 1] + (s1[i] != 'A');
    sux1[i] = s1[i] == 'A' ? sux1[i - 1] + 1 : 0;
  }
  for (int i = 1; i <= len2; i++) {
    pre2[i] = pre2[i - 1] + (s2[i] != 'A');
    sux2[i] = s2[i] == 'A' ? sux2[i - 1] + 1 : 0;
  }
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int l1, r1, l2, r2;
    scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
    if (Judge(l1, r1, l2, r2))
      ans[i] = 1;
    else
      ans[i] = 0;
  }
  for (int i = 0; i < q; i++) printf("%d", ans[i]);
  return 0;
}
