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 mod = (int)1e9 + 7;
const int inf = (1 << 30) - 1;
const int N = (int)5e5 + 5;
int n, m, q, bs[N], bt[N], lbs[N], lbt[N];
char t[N], second[N];
int main() {
  scanf("%s", second + 1);
  n = strlen(second + 1);
  scanf("%s", t + 1);
  m = strlen(t + 1);
  int bb = 0;
  for (int i = 1; i <= n; i++) {
    bs[i] = bs[i - 1] + (second[i] != 'A');
    if (second[i] != 'A') bb = i;
    lbs[i] = bb;
  }
  bb = 0;
  for (int i = 1; i <= m; i++) {
    bt[i] = bt[i - 1] + (t[i] != 'A');
    if (t[i] != 'A') bb = i;
    lbt[i] = bb;
  }
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int l, r, L, R;
    scanf("%d %d %d %d", &l, &r, &L, &R);
    int sa = r - max(lbs[r], l - 1);
    int ta = R - max(lbt[R], L - 1);
    if (sa < ta) {
      putchar('0');
      continue;
    }
    int sb = bs[r] - bs[l - 1];
    int tb = bt[R] - bt[L - 1];
    if (tb == sb) {
      if (sa >= ta && (sa - ta) % 3 == 0) {
        putchar('1');
        continue;
      }
    }
    if (sa != ta) sb += 2;
    if (sb == 0 && tb == 0) {
      putchar('1');
      continue;
    }
    if (sb > 0 && sb <= tb && ((sb + tb) & 1) == 0) {
      putchar('1');
    } else {
      putchar('0');
    }
  }
  puts("");
  return 0;
}
