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;
int sum[2][100005], trailing[2][100005];
int last[2][100005];
void solve() {
  string s[2];
  cin >> s[0] >> s[1];
  for (int i = 0; i < 2; ++i) {
    last[i][0] = -1;
    for (int j = 1; j <= s[i].size(); ++j) {
      int parity = s[i][j - 1] == 'A' ? 0 : 1;
      sum[i][j] = sum[i][j - 1] + parity;
      last[i][j] = last[i][j - 1];
      if (s[i][j - 1] == 'A') {
        trailing[i][j] = trailing[i][j - 1] + 1;
      } else {
        trailing[i][j] = 0;
      }
      if (parity) {
        last[i][j] = j;
      }
    }
  }
  int q;
  scanf("%d ", &q);
  int p = 1;
  while (q--) {
    int a, b, c, d;
    scanf("%d %d %d %d ", &a, &b, &c, &d);
    int parity = sum[0][b] - sum[0][a - 1];
    int parity2 = sum[1][d] - sum[1][c - 1];
    int len = min(b - a + 1, trailing[0][b]),
        len2 = min(d - c + 1, trailing[1][d]);
    if ((parity - parity2) % 2 != 0 || parity > parity2 || len < len2) {
      putchar('0');
    } else {
      bool isGood;
      if (parity == parity2) {
        isGood = (len - len2) % 3 == 0;
      } else {
        if (parity == 0) {
          if (len == len2) {
            isGood = false;
          } else {
            isGood = true;
          }
        } else {
          isGood = true;
        }
      }
      if (isGood) {
        putchar('1');
      } else {
        putchar('0');
      }
    }
    bool canConvert = true;
  }
  putchar('\n');
}
int main() {
  solve();
  return 0;
}
