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 MaxLen = 100000;
int f[2][MaxLen + 1][2];
char s[2][MaxLen];
int q;
void Get(int x) {
  int i, len = strlen(s[x]);
  for (i = 1; i <= len; i++) {
    char c = s[x][i - 1];
    f[x][i][0] = f[x][i - 1][0];
    f[x][i][1] = f[x][i - 1][1];
    if (c != 'A') {
      f[x][i][0]++;
      f[x][i][1] = 0;
    } else
      f[x][i][1]++;
  }
}
int main(int argc, char const *argv[]) {
  scanf("%s", s[0]);
  scanf("%s", s[1]);
  Get(0), Get(1);
  scanf("%d", &q);
  for (; q--;) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    int x = f[0][b][0] - f[0][a - 1][0], y = f[1][d][0] - f[1][c - 1][0],
        z = min(f[0][b][1], b - a + 1) - min(f[1][d][1], d - c + 1);
    if (x <= y)
      if ((x % 2) == (y % 2))
        if (z >= 0)
          if (x == y) {
            if ((z % 3) == 0) {
              printf("1");
              continue;
            }
          } else if (x || z) {
            printf("1");
            continue;
          }
    printf("0");
  }
  printf("\n");
  return 0;
}
