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 = 1e6 + 113;
const int INF = 1e9;
string s, t;
string getSimple(string pattern) {
  string ans = "";
  for (int i = 0; i < pattern.size(); ++i) {
    ans += (pattern[i] == 'A' ? '0' : '1');
  }
  return ans;
}
int sumS[N], sumT[N];
int suffS[N], suffT[N];
int main() {
  string p, q;
  cin >> p >> q;
  s = "0" + getSimple(p);
  t = "0" + getSimple(q);
  for (int i = 1; i < s.size(); ++i) {
    sumS[i] = sumS[i - 1] + s[i] - '0';
    suffS[i] = (s[i] == '1' ? 0 : suffS[i - 1] + 1);
  }
  for (int i = 1; i < t.size(); ++i) {
    sumT[i] = sumT[i - 1] + t[i] - '0';
    suffT[i] = (t[i] == '1' ? 0 : suffT[i - 1] + 1);
  }
  int t;
  cin >> t;
  for (int test = 0; test < t; ++test) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    int oneS = sumS[b] - sumS[a - 1];
    int oneT = sumT[d] - sumT[c - 1];
    int zeroS = min(b - a + 1, suffS[b]);
    int zeroT = min(d - c + 1, suffT[d]);
    if (oneT < oneS) {
      cout << 0;
      continue;
    }
    if ((oneT - oneS) % 2) {
      cout << 0;
      continue;
    }
    if (zeroT > zeroS) {
      cout << 0;
      continue;
    }
    if (oneT == oneS) {
      if ((zeroS - zeroT) % 3) {
        cout << 0;
      } else {
        cout << 1;
      }
      continue;
    }
    if (zeroT == zeroS) {
      if (oneS == 0) {
        cout << 0;
        continue;
      }
      cout << 1;
      continue;
    }
    cout << 1;
  }
}
