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;
string s, t;
int q;
int trailingS[maxn], trailingT[maxn], bS[maxn], bT[maxn], trailingCnt = 0,
                                                          bCnt = 0;
int main() {
  cin >> s >> t >> q;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == 'B' || s[i] == 'C') {
      trailingCnt = 0;
      bCnt++;
    } else {
      trailingCnt++;
    }
    trailingS[i + 1] = trailingCnt;
    bS[i + 1] = bCnt;
  }
  trailingCnt = 0, bCnt = 0;
  for (int i = 0; i < t.size(); i++) {
    if (t[i] == 'B' || t[i] == 'C') {
      trailingCnt = 0;
      bCnt++;
    } else {
      trailingCnt++;
    }
    trailingT[i + 1] = trailingCnt;
    bT[i + 1] = bCnt;
  }
  for (int i = 0; i < q; i++) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    int currentTrailingS = min(trailingS[b], b - a + 1),
        currentTrailingT = min(trailingT[d], d - c + 1);
    int currentbS = bS[b] - bS[a - 1], currentbT = bT[d] - bT[c - 1];
    if (currentbS == currentbT) {
      if (currentTrailingS - currentTrailingT >= 0 &&
          (currentTrailingS - currentTrailingT) % 3 == 0)
        printf("1");
      else
        printf("0");
    } else if (currentbS > 0 && currentbS < currentbT) {
      if ((currentbT - currentbS) % 2 == 0 &&
          currentTrailingS >= currentTrailingT)
        printf("1");
      else
        printf("0");
    } else if (currentbS == 0 && currentbT > 0) {
      if ((currentbT - currentbS) % 2 == 0 &&
          currentTrailingS > currentTrailingT)
        printf("1");
      else
        printf("0");
    } else
      printf("0");
  }
  return 0;
}
