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;
char s[100005], t[100005];
int mxs[100005], mxt[100005];
int bcs[100005], bct[100005];
int main() {
  scanf("%s%s", s + 1, t + 1);
  int ns = strlen(s + 1), nt = strlen(t + 1);
  for (int i = 1; i <= ns; i++) {
    if (s[i] == 'B' || s[i] == 'C') bcs[i]++, mxs[i] = i;
    bcs[i] += bcs[i - 1];
    mxs[i] = max(mxs[i], mxs[i - 1]);
  }
  for (int i = 1; i <= nt; i++) {
    if (t[i] == 'B' || t[i] == 'C') bct[i]++, mxt[i] = i;
    bct[i] += bct[i - 1];
    mxt[i] = max(mxt[i], mxt[i - 1]);
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    bool ans = 0;
    int ss = bcs[b] - bcs[a - 1];
    int st = bct[d] - bct[c - 1];
    int sa = min(b - a + 1, b - mxs[b]);
    int ta = min(d - c + 1, d - mxt[d]);
    if (ss > st)
      ans = 0;
    else if (ss == st) {
      if (ta > sa)
        ans = 0;
      else if ((sa - ta) % 3 == 0)
        ans = 1;
      else
        ans = 0;
    } else {
      int d = st - ss;
      if (d % 2 != 0)
        ans = 0;
      else {
        if (sa == ta && ss != 0)
          ans = 1;
        else if (sa <= ta)
          ans = 0;
        else
          ans = 1;
      }
    }
    printf("%d", ans);
  }
}
