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 + 1;
string s, t;
char c[N];
int cs[2][N], ct[2][N], q;
int main() {
  scanf(" %s", c);
  s = c;
  scanf(" %s", c);
  t = c;
  int at = 0;
  for (auto &i : s) {
    at++;
    if (i == 'C' || i == 'B') {
      i = 'B';
      cs[1][at] = cs[1][at - 1] + 1;
    } else {
      cs[1][at] = cs[1][at - 1];
      cs[0][at] = cs[0][at - 1] + 1;
    }
  }
  at = 0;
  for (auto &i : t) {
    at++;
    if (i == 'C' || i == 'B') {
      i = 'B';
      ct[1][at] = ct[1][at - 1] + 1;
    } else {
      ct[1][at] = ct[1][at - 1];
      ct[0][at] = ct[0][at - 1] + 1;
    }
  }
  scanf("%d", &q);
  for (int i = 0, a, b, c, d; i < q; i++) {
    scanf("%d%d%d%d", &a, &b, &c, &d);
    int difs = cs[1][b] - cs[1][a - 1];
    int dift = ct[1][d] - ct[1][c - 1];
    bool thereis = 0;
    int need = dift - difs;
    int trails = min(b - a + 1, cs[0][b]);
    int trailt = min(d - c + 1, ct[0][d]);
    if (need & 1) {
      printf("0");
      continue;
    }
    if (need < 0) {
      printf("0");
      continue;
    }
    if (!need) {
      if (trails < trailt || (trails - trailt) % 3) {
        printf("0");
        continue;
      }
      printf("1");
      continue;
    }
    if (need && trails > trailt) {
      trails = trailt;
      thereis = 1;
    }
    if ((trails - trailt) % 3 || trails < trailt) {
      printf("0");
      continue;
    }
    thereis |= difs;
    if (!thereis && need) {
      printf("0");
      continue;
    }
    printf("1");
  }
}
