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;
inline void read(int &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= f;
}
inline void judge() {
  freopen("in.txt", "r", stdin);
  freopen("out.txt", "w", stdout);
}
const int maxn = 100005;
char s[maxn], t[maxn];
int sum1[maxn], sum2[maxn], n, m, l[maxn], l2[maxn];
int main() {
  scanf("%s", s + 1);
  scanf("%s", t + 1);
  n = strlen(s + 1);
  m = strlen(t + 1);
  for (int i = (1); i <= (n); i++)
    sum1[i] = sum1[i - 1] + (s[i] == 'B' || s[i] == 'C');
  for (int i = (1); i <= (m); i++)
    sum2[i] = sum2[i - 1] + (t[i] == 'B' || t[i] == 'C');
  for (int i = (1); i <= (n); i++) {
    if (s[i] == 'A')
      l[i] = l[i - 1];
    else
      l[i] = i;
  }
  for (int i = (1); i <= (m); i++) {
    if (t[i] == 'A')
      l2[i] = l2[i - 1];
    else
      l2[i] = i;
  }
  int q;
  read(q);
  while (q--) {
    int a, b, c, d;
    read(a);
    read(b);
    read(c);
    read(d);
    int res1 = sum1[b] - sum1[a - 1], res2 = sum2[d] - sum2[c - 1];
    if ((res1 & 1) != (res2 & 1)) {
      putchar('0');
      continue;
    }
    if (res1 > res2) {
      putchar('0');
      continue;
    }
    int v = (b - max(a - 1, l[b]));
    int v2 = (d - max(c - 1, l2[d]));
    if (v < v2) {
      putchar('0');
      continue;
    }
    if (res1 < res2) {
      if (v == v2 && (res1 == 0))
        putchar('0');
      else
        putchar('1');
      continue;
    }
    if (res1 == res2) {
      if (v % 3 != v2 % 3) {
        putchar('0');
      } else {
        putchar('1');
      }
    }
  }
  return 0;
}
