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 INF32 = 1E9;
const long long INF64 = 4E18;
const long long M = 1E9 + 7;
const double EPS = 1E-9;
const double PI = 2 * acos(0);
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  string source, target;
  cin >> source >> target;
  int n1 = (int)source.size(), n2 = (int)target.size();
  source = ' ' + source;
  target = ' ' + target;
  for (int i = 1; i <= n1; ++i) {
    if (source[i] == 'C') source[i] = 'B';
  }
  for (int i = 1; i <= n2; ++i) {
    if (target[i] == 'C') target[i] = 'B';
  }
  vector<long long> prefix_b_source(n1 + 1, 0), trailing_a_source(n1 + 1, 0);
  vector<long long> prefix_b_target(n2 + 1, 0), trailing_a_target(n2 + 1, 0);
  for (int i = 1; i <= n1; ++i) {
    if (source[i] == 'B')
      prefix_b_source[i]++;
    else {
      trailing_a_source[i]++;
      if (source[i - 1] == 'A')
        trailing_a_source[i] += trailing_a_source[i - 1];
    }
    prefix_b_source[i] += prefix_b_source[i - 1];
  }
  for (int i = 1; i <= n2; ++i) {
    if (target[i] == 'B')
      prefix_b_target[i]++;
    else {
      trailing_a_target[i]++;
      if (target[i - 1] == 'A')
        trailing_a_target[i] += trailing_a_target[i - 1];
    }
    prefix_b_target[i] += prefix_b_target[i - 1];
  }
  int q;
  cin >> q;
  while (q--) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    int b_cnt_source = prefix_b_source[b] - prefix_b_source[a - 1];
    int b_cnt_target = prefix_b_target[d] - prefix_b_target[c - 1];
    int trailing_a_src = min(trailing_a_source[b], b - a + 1LL);
    int trailing_a_tgt = min(trailing_a_target[d], d - c + 1LL);
    if (trailing_a_src < trailing_a_tgt or b_cnt_source > b_cnt_target or
        (b_cnt_target - b_cnt_source) % 2 != 0) {
      cout << 0;
      continue;
    }
    if (b_cnt_source == 0) {
      if (b_cnt_target == 0) {
        if ((trailing_a_src - trailing_a_tgt) % 3 == 0)
          cout << 1;
        else
          cout << 0;
      } else {
        if (trailing_a_src <= trailing_a_tgt)
          cout << 0;
        else
          cout << 1;
      }
    } else {
      if (b_cnt_source == b_cnt_target) {
        if ((trailing_a_src - trailing_a_tgt) % 3 == 0)
          cout << 1;
        else
          cout << 0;
      } else if (b_cnt_source < b_cnt_target) {
        cout << 1;
      }
    }
  }
  return 0;
}
