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 long long INF = 1e18L;
const int inf = 0x3f3f3f3f;
const int maxn = 100005;
const int Mod = 1e9 + 7;
string S, T;
int tA[maxn], tB[maxn], sA[maxn], sB[maxn];
int q;
int ans[maxn];
int main() {
  ios::sync_with_stdio(false);
  cin >> S >> T;
  for (int i = 0; i < ((int)(S).size()); i++) {
    sB[i + 1] = sB[i];
    sA[i + 1] = 0;
    if (S[i] == 'A')
      sA[i + 1] = sA[i] + 1;
    else
      sB[i + 1] += 1;
  }
  for (int i = 0; i < ((int)(T).size()); i++) {
    tB[i + 1] = tB[i];
    tA[i + 1] = 0;
    if (T[i] == 'A')
      tA[i + 1] = tA[i] + 1;
    else
      tB[i + 1] += 1;
  }
  cin >> q;
  for (int i = 1; i <= q; i++) {
    int a, b, c, d, e = 0;
    cin >> a >> b >> c >> d;
    int s = min(b - a + 1, sA[b]), t = min(d - c + 1, tA[d]);
    if (s < t) {
      ans[i] = 0;
      continue;
    }
    if ((s - t) % 3 != 0) e = 2;
    int x = sB[b] - sB[a - 1], y = tB[d] - tB[c - 1];
    if (x == 0) {
      if (s == t) {
        if (y == 0)
          ans[i] = 1;
        else
          ans[i] = 0;
        continue;
      } else if (y == 0) {
        if (e != 0)
          ans[i] = 0;
        else
          ans[i] = 1;
        continue;
      }
    }
    x += e;
    if (x > y || x % 2 != y % 2)
      ans[i] = 0;
    else
      ans[i] = 1;
  }
  for (int i = 1; i <= q; i++) cout << ans[i];
  return 0;
}
