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 nmax = 1e5 + 42;
char inp[2][nmax];
pair<int, int> sum[2][nmax];
void get_inp(int id) {
  string now;
  cin >> now;
  int SZ = now.size();
  for (int i = 0; i < SZ; i++) {
    if (now[i] == 'C') now[i] = 'B';
    inp[id][i + 1] = now[i];
  }
  for (int i = 1; i <= SZ; i++) {
    sum[id][i] = sum[id][i - 1];
    if (inp[id][i] == 'A')
      sum[id][i].first++;
    else
      sum[id][i].second++;
  }
}
pair<int, int> find_sum(int id, int l, int r) {
  pair<int, int> ret = sum[id][r];
  ret.first = ret.first - sum[id][l - 1].first;
  ret.second = ret.second - sum[id][l - 1].second;
  return ret;
}
int a_ending(int id, int l, int r) {
  int ok = r + 1, not_ok = l - 1;
  while (ok - not_ok > 1) {
    int av = (ok + not_ok) / 2;
    if (find_sum(id, av, r).second == 0)
      ok = av;
    else
      not_ok = av;
  }
  return r - ok + 1;
}
bool query() {
  int l, r, lq, rq;
  cin >> l >> r >> lq >> rq;
  int bs = find_sum(0, l, r).second, bt = find_sum(1, lq, rq).second;
  int a_end_s = a_ending(0, l, r), a_end_t = a_ending(1, lq, rq);
  if (bs % 2 != bt % 2) return 0;
  if (bs > bt) return 0;
  if (bs == bt) return a_end_s >= a_end_t && a_end_s % 3 == a_end_t % 3;
  if (bs && bs < bt) return a_end_s >= a_end_t;
  return a_end_s > a_end_t;
}
int main() {
  get_inp(0);
  get_inp(1);
  int q;
  cin >> q;
  for (int i = 1; i <= q; i++) cout << query();
  return 0;
}
