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 N = (int)1e6 + 7;
const int MOD = (int)1e9 + 7;
string a, b;
int n;
int alen;
int blen;
vector<int> listA, listB;
int sa[N];
int sb[N];
int getA(int l, int r) {
  int pos = upper_bound(listA.begin(), listA.end(), r) - listA.begin();
  if (pos == 0) return l - 1;
  pos--;
  return listA[pos];
}
int getB(int l, int r) {
  int pos = upper_bound(listB.begin(), listB.end(), r) - listB.begin();
  if (pos == 0) return l - 1;
  pos--;
  return listB[pos];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> a >> b;
  alen = a.size();
  blen = b.size();
  a = "#" + a;
  b = "#" + b;
  for (int i = 1; i <= alen; i++) {
    if (a[i] != 'A') listA.push_back(i);
  }
  for (int i = 1; i <= blen; i++) {
    if (b[i] != 'A') listB.push_back(i);
  }
  for (int i = 1; i <= alen; i++) {
    sa[i] = sa[i - 1];
    if (a[i] != 'A') ++sa[i];
  }
  for (int i = 1; i <= blen; i++) {
    sb[i] = sb[i - 1];
    if (b[i] != 'A') ++sb[i];
  }
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int al, ar;
    int bl, br;
    cin >> al >> ar >> bl >> br;
    int o1 = sa[ar] - sa[al - 1];
    int o2 = sb[br] - sb[bl - 1];
    int z1 = ar - max(al - 1, getA(al, ar));
    int z2 = br - max(bl - 1, getB(bl, br));
    bool good = 0;
    if (o1 <= o2 && o1 % 2 == o2 % 2) {
      int need = (o2 - o1) / 2;
      if (need == 0) {
        good = (z1 >= z2 && z1 % 3 == z2 % 3);
      } else {
        if (o1) {
          if (z1 >= z2) {
            good = 1;
          }
        } else {
          if (z1 > z2) {
            good = 1;
          }
        }
      }
    }
    cout << good;
  }
  return 0;
}
