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;
#pragma comment(linker, "/STACK:64000000")
const int infi = 1e9 + 7;
const long long infl = 2e18 + 7;
string S, T;
int sum1[100500];
int sum2[100500];
int cnt1[100500], cnt2[100500];
void build(string &s, int *sum, int *cnt) {
  int n = s.size();
  int cnt1 = 0;
  for (int i = 0; i < (n); ++i) {
    if (!s[i])
      sum[i] = 1;
    else
      sum[i] = 0;
    if (i) sum[i] += sum[i - 1];
    if (s[i]) {
      ++cnt1;
    } else
      cnt1 = 0;
    cnt[i] = cnt1;
  }
  return;
}
int get(int *sum, int l, int r) {
  if (l) return sum[r] - sum[l - 1];
  return sum[r];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> S >> T;
  for (int i = 0; i < ((int)S.size()); ++i)
    if (S[i] == 'A')
      S[i] = 1;
    else
      S[i] = 0;
  for (int i = 0; i < ((int)T.size()); ++i)
    if (T[i] == 'A')
      T[i] = 1;
    else
      T[i] = 0;
  build(S, sum1, cnt1);
  build(T, sum2, cnt2);
  int m;
  cin >> m;
  for (int i = 0; i < (m); ++i) {
    int l1, r1, l2, r2;
    cin >> l1 >> r1 >> l2 >> r2;
    --l1, --r1, --l2, --r2;
    int a = get(sum1, l1, r1), b = get(sum2, l2, r2);
    int c1 = cnt1[r1], c2 = cnt2[r2];
    c1 = min(c1, r1 - l1 + 1);
    c2 = min(c2, r2 - l2 + 1);
    if (a % 2 != b % 2 || a > b) {
      cout << 0;
      continue;
    }
    if (S[r1] == 1) {
      if (T[r2] == 1) {
        if (c1 < c2) {
          cout << 0;
          continue;
        }
        if (c1 % 3 != c2 % 3) {
          a += 2;
          if (a > b) {
            cout << 0;
            continue;
          }
          cout << 1;
          continue;
        }
        if (a != 0) {
          cout << 1;
          continue;
        }
        if (b == 0) {
          cout << 1;
          continue;
        }
        if (c2 == c1) {
          cout << 0;
          continue;
        }
        a += 2;
        if (a > b)
          cout << 0;
        else
          cout << 1;
      } else {
        if (a < b) {
          cout << 1;
          continue;
        }
        if (c1 % 3) {
          cout << 0;
          continue;
        }
        if (r1 - l1 + 1 == c1) {
          cout << 0;
          assert(0);
        } else
          cout << 1;
      }
    } else {
      if (T[r2] == 1) {
        cout << 0;
      } else {
        cout << 1;
      }
    }
  }
  return 0;
}
