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;
string s, s1;
int pref[200000], pref1[200000], suf[200000], suf1[200000];
int main() {
  cin >> s;
  cin >> s1;
  int kek = 0;
  if (s[0] == 'A') {
    pref[0] = 0;
    suf[0] = 1;
    kek = 1;
  } else
    pref[0] = 1;
  for (int i = 1; i < s.size(); ++i) {
    if (!(s[i] == 'A')) {
      pref[i] = pref[i - 1] + 1;
      kek = 0;
    } else {
      ++kek;
      pref[i] = pref[i - 1];
    }
    suf[i] = kek;
  }
  kek = 0;
  if (s1[0] == 'A') {
    pref1[0] = 0;
    suf1[0] = 1;
    kek = 1;
  } else
    pref1[0] = 1;
  for (int i = 1; i < s1.size(); ++i) {
    if (!(s1[i] == 'A')) {
      pref1[i] = pref1[i - 1] + 1;
      kek = 0;
    } else {
      pref1[i] = pref1[i - 1];
      ++kek;
    }
    suf1[i] = kek;
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    int l, r, l1, r1;
    cin >> l >> r >> l1 >> r1;
    --l;
    --r;
    --l1;
    --r1;
    int lol, lol1;
    if (l == 0)
      lol = 0;
    else
      lol = pref[l - 1];
    if (l1 == 0)
      lol1 = 0;
    else
      lol1 = pref1[l1 - 1];
    int b1 = pref[r] - lol;
    int b2 = pref1[r1] - lol1;
    int a1 = suf[r];
    int a2 = suf1[r1];
    if (b1 == 0) a1 = r - l + 1;
    if (b2 == 0) a2 = r1 - l1 + 1;
    if ((b1 == 0) && (b2 > 0)) --a1;
    if ((b1 <= b2) && (a1 >= a2) && (b1 % 2 == b2 % 2)) {
      if ((a1 % 3 == a2 % 3))
        cout << 1;
      else if (b1 + 2 <= b2)
        cout << 1;
      else
        cout << 0;
    } else
      cout << 0;
  }
  cout << "\n";
}
