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 MOD = 1e9 + 7;
const long double E = 1e-9;
unsigned char ccc;
bool _minus = false;
template <typename T>
inline T sqr(T t) {
  return (t * t);
}
inline void read(long long &n) {
  n = 0;
  _minus = false;
  while (true) {
    ccc = getchar();
    if (ccc == ' ' || ccc == '\n') break;
    if (ccc == '-') {
      _minus = true;
      continue;
    }
    n = n * 10 + ccc - '0';
  }
  if (_minus) n *= -1;
}
inline bool read(int &n) {
  n = 0;
  _minus = false;
  while (true) {
    ccc = getchar();
    if (ccc == ' ' || ccc == '\n') {
      if (ccc == '\n') return true;
      break;
    }
    if (ccc == '-') {
      _minus = true;
      continue;
    }
    n = n * 10 + ccc - '0';
  }
  if (_minus) n *= -1;
  return false;
}
char wwww[19];
int kkkk;
inline void write(long long y) {
  long long x = y;
  kkkk = 0;
  if (x < 0) {
    putchar('-');
    x *= -1;
  }
  if (!x) {
    ++kkkk;
    wwww[kkkk] = '0';
  } else
    while (x) {
      ++kkkk;
      wwww[kkkk] = char(x % 10 + '0');
      x /= 10;
    }
  for (int i = kkkk; i >= 1; --i) putchar(wwww[i]);
}
const int MAX = 1e5 + 10;
int n, m;
char a[MAX], b[MAX];
int last_b_1[MAX], last_b_2[MAX], how_many_b_1[MAX], how_many_b_2[MAX];
int solve(int l1, int r1, int l2, int r2) {
  int b1 = how_many_b_1[r1] - how_many_b_1[l1 - 1];
  int b2 = how_many_b_2[r2] - how_many_b_2[l2 - 1];
  int sz1 = r1 - l1 + 1;
  int sz2 = r2 - l2 + 1;
  if (b1 > b2) {
    return false;
  }
  if ((b2 - b1) % 2 == 1) {
    return false;
  }
  if (b1 == 0 && b2 == 0) {
    return (sz1 >= sz2 && (sz1 - sz2) % 3 == 0);
  }
  int a2 = r2 - last_b_2[r2];
  if (b1 == 0) {
    return (sz1 > a2);
  }
  int a1 = r1 - last_b_1[r1];
  if (a1 < a2) {
    return false;
  }
  if (a1 > a2) {
    if (b1 == b2) {
      if ((a1 - a2) % 3 != 0) {
        return false;
      }
    }
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  srand((unsigned int)time(NULL));
  cout.precision(10);
  cout << fixed;
  string s;
  cin >> s;
  n = (int)s.size();
  for (int i = 0; i < n; i++) {
    a[i + 1] = s[i];
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] == 'C') {
      a[i] = 'B';
    }
    last_b_1[i] = (a[i] == 'B' ? i : last_b_1[i - 1]);
    how_many_b_1[i] = how_many_b_1[i - 1] + (a[i] == 'B' ? 1 : 0);
  }
  cin >> s;
  m = (int)s.size();
  for (int i = 0; i < m; i++) {
    b[i + 1] = s[i];
  }
  for (int i = 1; i <= m; i++) {
    if (b[i] == 'C') {
      b[i] = 'B';
    }
    last_b_2[i] = (b[i] == 'B' ? i : last_b_2[i - 1]);
    how_many_b_2[i] = how_many_b_2[i - 1] + (b[i] == 'B' ? 1 : 0);
  }
  int q;
  cin >> q;
  while (q--) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    cout << solve(a, b, c, d);
  }
  cout << "\n";
}
