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 + 10;
int s[N][3], s1[N][3], n, m;
int a[N], a1[N];
int main() {
  char c;
  for (c = getchar(); c < 'A' || c > 'C'; c = getchar())
    ;
  for (; c >= 'A' && c <= 'C'; c = getchar()) {
    int x = c - 'A';
    n++;
    memcpy(s[n], s[n - 1], sizeof(s[n]));
    s[n][x]++;
    if (x == 0) a[n] = a[n - 1] + 1;
  }
  for (c = getchar(); c < 'A' || c > 'C'; c = getchar())
    ;
  for (; c >= 'A' && c <= 'C'; c = getchar()) {
    int x = c - 'A';
    m++;
    memcpy(s1[m], s1[m - 1], sizeof(s1[m]));
    s1[m][x]++;
    if (x == 0) a1[m] = a1[m - 1] + 1;
  }
  int q;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int l1, l2, r1, r2;
    scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
    int x = s[r1][2] - s[l1 - 1][2] + s[r1][1] - s[l1 - 1][1];
    int y = s1[r2][2] - s1[l2 - 1][2] + s1[r2][1] - s1[l2 - 1][1];
    int ax = min(a[r1], r1 - l1 + 1), ay = min(a1[r2], r2 - l2 + 1);
    int c = ax - ay;
    if (x <= y) {
      if (x % 2 == y % 2) {
        if (c >= 0) {
          if (x == y) {
            if (c % 3 == 0) {
              printf("1");
              continue;
            }
          } else {
            if (x || c) {
              printf("1");
              continue;
            }
          }
        }
      }
    }
    printf("0");
  }
  return 0;
}
