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 = 1e5 + 10;
char S[maxn], T[maxn];
int sx, sy, tx, ty;
int sums[maxn], lasts[maxn], sumt[maxn], lastt[maxn];
bool Judge(int del, int x, int y, int op) {
  if (sx > tx) return 0;
  if (del & 1) return 0;
  if (x < y) return 0;
  if (!del) return (x - y) % 3 == 0;
  if (op) {
    for (int i = 1; i <= 3 && i <= x; i++)
      if (Judge(del - 2, x - i, y, 0)) return 1;
    return 0;
  }
  return 1;
}
int main() {
  scanf("%s\n%s\n", S + 1, T + 1);
  int len1 = strlen(S + 1);
  int len2 = strlen(T + 1);
  for (int i = 1; i <= len1; i++) {
    sums[i] = sums[i - 1] + (S[i] != 'A');
    lasts[i] = S[i] == 'A' ? lasts[i - 1] + 1 : 0;
  }
  for (int i = 1; i <= len2; i++) {
    sumt[i] = sumt[i - 1] + (T[i] != 'A');
    lastt[i] = T[i] == 'A' ? lastt[i - 1] + 1 : 0;
  }
  int Q;
  scanf("%d", &Q);
  for (int i = 1, a, b, c, d; i <= Q; i++) {
    scanf("%d%d%d%d", &a, &b, &c, &d);
    sx = sums[b] - sums[a - 1];
    sy = min(lasts[b], b - a + 1);
    tx = sumt[d] - sumt[c - 1];
    ty = min(lastt[d], d - c + 1);
    bool ok;
    ok = Judge(tx - sx, sy, ty, lasts[b] > b - a);
    printf("%d", ok);
  }
  return 0;
}
