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 long long MOD = 1e9 + 7;
const int MAXN = 1e6 + 9;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-6;
inline int sgn(double x) {
  if (x < -EPS) return -1;
  return x > EPS;
}
inline int cmp(double x, double y) { return sgn(x - y); }
int n, m;
char a[MAXN], b[MAXN];
int num1[MAXN], num2[MAXN];
int pre1[MAXN], pre2[MAXN];
void solve() {
  int l1, l2, r1, r2;
  while (m--) {
    scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
    int zero1 = pre1[r1];
    int zero2 = pre2[r2];
    int lastone1 = zero1 < l1 ? r1 - l1 + 1 : r1 - zero1;
    int lastone2 = zero2 < l2 ? r2 - l2 + 1 : r2 - zero2;
    if (lastone1 < lastone2) {
      printf("0");
      continue;
    }
    lastone1 -= lastone2;
    int mustadd = 0;
    int canadd = 0;
    if (lastone1 % 3) {
      mustadd = 1;
    } else if (lastone1) {
      canadd = 1;
    }
    int rr1 = max(zero1, l1);
    int rr2 = max(zero2, l2);
    int qianzero1 = num1[rr1] - num1[l1 - 1] + mustadd * 2;
    int qianzero2 = num2[rr2] - num2[l2 - 1];
    if (canadd && qianzero1 == 0 && qianzero2) qianzero1 += 2;
    if (qianzero1 == 0 && qianzero2) {
      printf("0");
      continue;
    }
    if ((qianzero1) % 2 == (qianzero2) % 2 && (qianzero1 <= qianzero2))
      printf("1");
    else
      printf("0");
  }
  printf("\n");
}
void init() {
  scanf("%s", a + 1);
  n = strlen(a + 1);
  for (int i = (int)(1); i <= (int)(n); i++) {
    num1[i] = num1[i - 1];
    pre1[i] = pre1[i - 1];
    if (a[i] != 'A') {
      num1[i] += 1;
      pre1[i] = i;
    }
  }
  scanf("%s", b + 1);
  n = strlen(b + 1);
  for (int i = (int)(1); i <= (int)(n); i++) {
    num2[i] = num2[i - 1];
    pre2[i] = pre2[i - 1];
    if (b[i] != 'A') {
      num2[i] += 1;
      pre2[i] = i;
    }
  }
  scanf("%d", &m);
}
int main() {
  int T = 1;
  while (T--) {
    init();
    solve();
  }
  return 0;
}
