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;
int sufs[100010], suft[100010];
int nums[100010], numt[100010];
char s[100010], t[100010];
int main() {
  cin >> s >> t;
  int n = strlen(s), m = strlen(t), q, a, b, c, d;
  for (int i = 0; i < n; i++)
    if (s[i] == 'A')
      sufs[i] = sufs[i - 1] + 1, nums[i] = nums[i - 1];
    else
      nums[i] = nums[i - 1] + 1;
  for (int i = 0; i < m; i++)
    if (t[i] == 'A')
      suft[i] = suft[i - 1] + 1, numt[i] = numt[i - 1];
    else
      numt[i] = numt[i - 1] + 1;
  cin >> q;
  for (int i = 1; i <= q; i++) {
    scanf("%d%d%d%d", &a, &b, &c, &d), a--, b--, c--, d--;
    int ss = min(sufs[b], b - a + 1), st = min(suft[d], d - c + 1);
    int t1 = nums[b] - nums[a - 1], t2 = numt[d] - numt[c - 1];
    if (ss == b - a + 1 && st != d - c + 1) {
      ss--, t1 += 2;
      if (ss > st) ss = st;
    }
    if (ss < st)
      putchar('0');
    else {
      if ((ss - st) % 3) t1 += 2, ss = st;
      if (t1 <= t2 && (t1 - t2) % 2 == 0)
        putchar('1');
      else
        putchar('0');
    }
  }
  return 0;
}
