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 MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1);
const double EPS = 1e-10;
const int N = 100010;
int n, m;
string s, t;
int sCntA[N], tCntA[N];
int sSumB[N], tSumB[N];
bool canTransit() {
  int l1, r1, l2, r2;
  cin >> l1 >> r1 >> l2 >> r2;
  --l1, --l2;
  int sb = sSumB[r1] - sSumB[l1];
  int tb = tSumB[r2] - tSumB[l2];
  if ((sb + tb) & 1) {
    return false;
  }
  if (sb > tb) {
    return false;
  }
  int sa = min(r1 - l1, sCntA[r1 - 1]);
  int ta = min(r2 - l2, tCntA[r2 - 1]);
  if (sa < ta) {
    return false;
  }
  if (tb == sb) {
    return sa % 3 == ta % 3;
  }
  if (sb == 0 && sa == ta) {
    return false;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> s >> t;
  n = (int)s.size();
  m = (int)t.size();
  for (int i = 0; i < n; ++i) {
    sSumB[i + 1] = sSumB[i];
    if (s[i] == 'A') {
      sCntA[i] = (i > 0 ? sCntA[i - 1] : 0) + 1;
    } else {
      ++sSumB[i + 1];
    }
  }
  for (int i = 0; i < m; ++i) {
    tSumB[i + 1] = tSumB[i];
    if (t[i] == 'A') {
      tCntA[i] = (i > 0 ? tCntA[i - 1] : 0) + 1;
    } else {
      ++tSumB[i + 1];
    }
  }
  int q;
  cin >> q;
  while (q--) {
    cout << canTransit();
  }
}
