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 maxn = 1e5 + 10;
int prefA[2][maxn], prefB[2][maxn];
vector<int> posB[2];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  string s, t;
  cin >> s >> t;
  for (int i = 0; i < s.size(); ++i) {
    prefA[0][i + 1] = prefA[0][i];
    prefB[0][i + 1] = prefB[0][i];
    if (s[i] == 'A')
      prefA[0][i + 1]++;
    else
      prefB[0][i + 1]++, posB[0].push_back(i + 1);
  }
  for (int i = 0; i < t.size(); ++i) {
    prefA[1][i + 1] = prefA[1][i];
    prefB[1][i + 1] = prefB[1][i];
    if (t[i] == 'A')
      prefA[1][i + 1]++;
    else
      prefB[1][i + 1]++, posB[1].push_back(i + 1);
  }
  int q;
  cin >> q;
  while (q--) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    int cntB[2];
    cntB[0] = prefB[0][b] - prefB[0][a - 1];
    cntB[1] = prefB[1][d] - prefB[1][c - 1];
    if (cntB[0] > cntB[1] || ((cntB[1] - cntB[0]) & 1))
      cout << 0;
    else {
      int lastA[2] = {};
      int L = -1, R = posB[0].size();
      while (L + 1 < R) {
        int M = (L + R) / 2;
        if (posB[0][M] <= b)
          L = M;
        else
          R = M;
      }
      int pos;
      if (L == -1)
        pos = a - 1;
      else
        pos = max(posB[0][L], a - 1);
      lastA[0] = prefA[0][b] - prefA[0][pos];
      L = -1;
      R = posB[1].size();
      while (L + 1 < R) {
        int M = (L + R) / 2;
        if (posB[1][M] <= d)
          L = M;
        else
          R = M;
      }
      if (L == -1)
        pos = c - 1;
      else
        pos = max(posB[1][L], c - 1);
      lastA[1] = prefA[1][d] - prefA[1][pos];
      if (lastA[0] < lastA[1]) {
        cout << 0;
        continue;
      }
      if (cntB[0] == cntB[1]) {
        if ((lastA[0] - lastA[1]) % 3 == 0)
          cout << 1;
        else
          cout << 0;
      } else if (cntB[0] == 0) {
        if (lastA[0] != lastA[1])
          cout << 1;
        else
          cout << 0;
      } else {
        cout << 1;
      }
    }
  }
  return 0;
}
