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 = 100010;
char s[MAXN], t[MAXN];
int cnts[MAXN], cntt[MAXN], coss[MAXN], cost[MAXN];
int main() {
  scanf("%s%s", s + 1, t + 1);
  for (int i = 1; s[i]; i++) {
    cnts[i] = cnts[i - 1] + (s[i] == 'A');
    if (s[i] == 'A')
      coss[i] = coss[i - 1] + 1;
    else
      coss[i] = 0;
  }
  for (int i = 1; t[i]; i++) {
    cntt[i] = cntt[i - 1] + (t[i] == 'A');
    if (t[i] == 'A')
      cost[i] = cost[i - 1] + 1;
    else
      cost[i] = 0;
  }
  int q;
  scanf("%d", &q);
  int l1, r1, l2, r2;
  for (int i = 1; i <= q; i++) {
    scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
    int ln1 = r1 - l1 + 1 - cnts[r1] + cnts[l1 - 1];
    int ln2 = r2 - l2 + 1 - cntt[r2] + cntt[l2 - 1];
    int x = min(r1 - l1 + 1, coss[r1]), y = min(r2 - l2 + 1, cost[r2]);
    if (x == r1 - l1 + 1 && y < r2 - l2 + 1) {
      if (x > y)
        x = y;
      else
        x--;
      ln1 += 2;
    }
    if (x < y) {
      printf("0");
      continue;
    }
    if ((x - y) % 3 != 0) ln1 += 2;
    if (ln1 <= ln2 && (ln1 - ln2) % 2 == 0)
      printf("1");
    else
      printf("0");
  }
  printf("\n");
}
