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 N = 1E5 + 5;
int s[N][2], pre[N][2], cnt[N][2], len[2], T, l, r, t, w;
void Init() {
  for (int i = 0; i <= 1; i++) {
    char c = getchar();
    while ('A' <= c && c <= 'C') {
      s[++len[i]][i] = c - 'A';
      pre[len[i]][i] = pre[len[i] - 1][i];
      if (c - 'A') {
        pre[len[i]][i]++;
      }
      c = getchar();
    }
    scanf("\n");
  }
  for (int i = 0; i <= 1; i++) {
    int last = 0;
    for (int j = 1; j <= len[i]; j++) {
      if (s[j][i]) last = j;
      cnt[j][i] = j - last;
    }
  }
}
void Work() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d%d%d%d", &l, &r, &t, &w);
    int tmp1 = pre[r][0] - pre[l - 1][0];
    int tmp2 = pre[w][1] - pre[t - 1][1];
    if (tmp1 > tmp2) {
      printf("0");
      continue;
    }
    if (tmp1 % 2 != tmp2 % 2) {
      printf("0");
      continue;
    }
    int cnt1 = cnt[r][0];
    cnt1 = min(cnt1, r - l + 1);
    int cnt2 = cnt[w][1];
    cnt2 = min(cnt2, w - t + 1);
    if (cnt1 < cnt2) {
      printf("0");
      continue;
    }
    if (tmp1 == tmp2 && cnt1 % 3 != cnt2 % 3) {
      printf("0");
      continue;
    }
    if (tmp1 != tmp2 && tmp1 == 0 && cnt1 == cnt2) {
      printf("0");
      continue;
    }
    printf("1");
  }
}
int main() {
  Init();
  Work();
  return 0;
}
