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 maxn = 101010;
int Bs[maxn] = {0};
int Bt[maxn] = {0};
int As[maxn] = {0};
int At[maxn] = {0};
int AAs[maxn] = {0};
int AAt[maxn] = {0};
int main() {
  cin.tie(NULL);
  ios::sync_with_stdio(false);
  string s, t;
  cin >> s >> t;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == 'B' || s[i] == 'C') {
      Bs[i + 1] = 1;
    }
    Bs[i + 1] += Bs[i];
  }
  for (int i = 0; i < t.size(); i++) {
    if (t[i] == 'B' || t[i] == 'C') {
      Bt[i + 1] = 1;
    }
    Bt[i + 1] += Bt[i];
  }
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == 'A') {
      AAs[i + 1] = 1;
    }
    AAs[i + 1] += AAs[i];
  }
  for (int i = 0; i < t.size(); i++) {
    if (t[i] == 'A') {
      AAt[i + 1] = 1;
    }
    AAt[i + 1] += AAt[i];
  }
  int k = 0;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == 'A') {
      k++;
    } else {
      k = 0;
    }
    As[i + 1] = k;
  }
  for (int i = 0; i < t.size(); i++) {
    if (t[i] == 'A') {
      k++;
    } else {
      k = 0;
    }
    At[i + 1] = k;
  }
  int test;
  cin >> test;
  for (int id = 0; id < test; id++) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    if (Bs[b] - Bs[a - 1] > Bt[d] - Bt[c - 1] ||
        (Bt[d] - Bt[c - 1] - (Bs[b] - Bs[a - 1])) % 2 != 0) {
      cout << 0;
      continue;
    }
    if (min(As[b], AAs[b] - AAs[a - 1]) < min(At[d], AAt[d] - AAt[c - 1])) {
      cout << 0;
      continue;
    }
    if (Bs[b] - Bs[a - 1] == 0 && Bt[d] - Bt[c - 1] > 0) {
      if (min(As[b], AAs[b] - AAs[a - 1]) <= min(At[d], AAt[d] - AAt[c - 1])) {
        cout << 0;
        continue;
      }
    }
    if (Bs[b] - Bs[a - 1] == Bt[d] - Bt[c - 1]) {
      if (min(As[b], AAs[b] - AAs[a - 1]) % 3 !=
          min(At[d], AAt[d] - AAt[c - 1]) % 3) {
        cout << 0;
        continue;
      }
    }
    cout << 1;
  }
  return 0;
}
