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>
#pragma comment(linker, "/STACK:536870912")
#pragma warning(disable : 4996)
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int N = 1e5 + 100;
char s[N], t[N];
int ps[N], pt[N], es[N], et[N];
int main() {
  scanf(" %s %s", s, t);
  int ls = strlen(s), lt = strlen(t);
  int last = -1;
  for (int i = 0; i < ls; i++) {
    ps[i + 1] = ps[i];
    if (s[i] != 'A') {
      ps[i + 1]++;
      last = i;
    } else
      es[i + 1] = i - last;
  }
  last = -1;
  for (int i = 0; i < lt; i++) {
    pt[i + 1] = pt[i];
    if (t[i] != 'A') {
      pt[i + 1]++;
      last = i;
    } else
      et[i + 1] = i - last;
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    a--;
    c--;
    int ss = min(es[b], b - a), tt = min(et[d], d - c), x = ps[b] - ps[a],
        y = pt[d] - pt[c];
    if (x % 2 != y % 2 || x > y || ss < tt || x == 0 && y > 0 && ss - 1 < tt) {
      printf("0");
    } else if (ss % 3 == tt % 3 || x + 2 <= y) {
      printf("1");
    } else
      printf("0");
  }
  return 0;
}
