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;
int BS[100005];
int BT[100005];
char S[100005];
char T[100005];
int SN, TN;
int ans[100005];
int AS[100005];
int AT[100005];
int main() {
  scanf(" %s", S + 1);
  scanf(" %s", T + 1);
  SN = (int)strlen(S + 1);
  TN = (int)strlen(T + 1);
  S[0] = T[0] = 0;
  for (int i = 1; i <= SN; i++) {
    BS[i] = BS[i - 1];
    if (S[i] == 'B' || S[i] == 'C') BS[i]++;
  }
  for (int i = 1; i <= TN; i++) {
    BT[i] = BT[i - 1];
    if (T[i] == 'B' || T[i] == 'C') BT[i]++;
  }
  for (int i = 1; i <= SN; i++) {
    if (S[i] != 'A')
      AS[i] = 0;
    else
      AS[i] = AS[i - 1] + 1;
  }
  for (int i = 1; i <= TN; i++) {
    if (T[i] != 'A')
      AT[i] = 0;
    else
      AT[i] = AT[i - 1] + 1;
  }
  int Q;
  scanf("%d", &Q);
  for (int q = 0; q < Q; q++) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    int scnt = BS[b] - BS[a - 1];
    int tcnt = BT[d] - BT[c - 1];
    if (scnt > tcnt) {
      ans[q] = 0;
      continue;
    }
    if (scnt % 2 != tcnt % 2) {
      ans[q] = 0;
      continue;
    }
    int x = min(b - a + 1, AS[b]), y = min(d - c + 1, AT[d]);
    if (x < y) {
      ans[q] = 0;
      continue;
    }
    if (x > y && (x - y) % 3 && scnt + 2 > tcnt) {
      ans[q] = 0;
      continue;
    }
    if (x == y && x == b - a + 1 && y != d - c + 1) {
      ans[q] = 0;
      continue;
    }
    ans[q] = 1;
  }
  for (int i = 0; i < Q; i++) printf("%d", ans[i]);
  return 0;
}
