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;
string a, b;
const int maxn = 100005;
int sufa1[maxn], sufb1[maxn], sufa2[maxn], sufb2[maxn];
int main() {
  cin >> a >> b;
  int m;
  for (int i = 0; a[i] != '\0'; i++) {
    if (a[i] == 'A') {
      sufa1[i + 1] = sufa1[i] + 1;
      sufb1[i + 1] = sufb1[i];
    } else {
      sufa1[i + 1] = 0;
      sufb1[i + 1] = sufb1[i] + 1;
    }
  }
  for (int i = 0; b[i] != '\0'; i++) {
    if (b[i] == 'A') {
      sufa2[i + 1] = sufa2[i] + 1;
      sufb2[i + 1] = sufb2[i];
    } else {
      sufa2[i + 1] = 0;
      sufb2[i + 1] = sufb2[i] + 1;
    }
  }
  scanf("%d", &m);
  while (m--) {
    int l, r, l2, r2;
    scanf("%d%d%d%d", &l, &r, &l2, &r2);
    int a1, b1, a2, b2;
    if (r - l + 1 < sufa1[r])
      a1 = r - l + 1;
    else
      a1 = sufa1[r];
    b1 = sufb1[r] - sufb1[l - 1];
    if (r2 - l2 + 1 < sufa2[r2])
      a2 = r2 - l2 + 1;
    else
      a2 = sufa2[r2];
    b2 = sufb2[r2] - sufb2[l2 - 1];
    if (a1 < a2) {
      printf("0");
      continue;
    }
    if ((a1 % 3) != (a2 % 3)) b1 += 2;
    if (b1 == 0 && b2 > 0 && a1 > a2) {
      b1 += 2;
    }
    if (b1 == 0 && b2 > 0) {
      printf("0");
      continue;
    }
    if (b1 <= b2 && (b2 - b1) % 2 == 0) {
      printf("1");
    } else {
      printf("0");
    }
  }
  return 0;
}
