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;
using ll = long long;
const int maxLen = 2e5 + 5;
int totLen, toAdd;
string str;
int semiOpen[maxLen];
vector<int> posStar;
pair<int, int> getInfo(int lo, int ri) {
  int nbStar = semiOpen[ri + 1] - semiOpen[lo];
  int nbEnd = ri - lo + 1;
  if (nbStar > 0) {
    auto it = upper_bound(posStar.begin(), posStar.end(), ri);
    --it;
    nbEnd = ri - (*it);
  }
  return {nbStar, nbEnd};
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  if (true) {
    string s, t;
    cin >> s >> t;
    str = s + t;
    totLen = str.size();
    toAdd = s.size();
  }
  for (int i = 0; i < totLen; ++i) {
    int isStar = (str[i] != 'A');
    semiOpen[i + 1] = semiOpen[i] + isStar;
    if (isStar) posStar.push_back(i);
  }
  int nbQuery;
  cin >> nbQuery;
  for (int iQuery = 0; iQuery < nbQuery; ++iQuery) {
    int l1, r1, l2, r2;
    cin >> l1 >> r1 >> l2 >> r2;
    --l1;
    --r1;
    --l2;
    --r2;
    l2 += toAdd;
    r2 += toAdd;
    auto g1 = getInfo(l1, r1), g2 = getInfo(l2, r2);
    int cond = g1.first < g2.first;
    if (g1.first == g2.first) {
      cond = g1.second % 3 == g2.second % 3;
    }
    cond &= (g1.first <= g2.first);
    cond &= (g1.first % 2 == g2.first % 2);
    cond &= g1.second >= g2.second;
    if (g1.first == 0) {
      bool isEqual = (r2 - l2 + 1) == (r1 - l1 + 1);
      isEqual &= (g1 == g2);
      cond &= (g1.second > g2.second || isEqual);
    }
    cout << cond;
  }
  cout << "\n";
}
