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, mod = 1e9 + 7;
char s[maxn], t[maxn];
int bita[maxn], bitb[maxn], ans[maxn], pa[maxn], pb[maxn];
int lowbit(int x) { return x & -x; }
void update(int b[], int v, int val, int up) {
  while (v <= up) {
    b[v] += val;
    v += lowbit(v);
  }
}
int sum(int b[], int v) {
  int res = 0;
  while (v) {
    res += b[v];
    v -= lowbit(v);
  }
  return res;
}
void init() {
  int n = strlen(s + 1);
  int m = strlen(t + 1);
  for (int i = 1; i <= n; i++) {
    update(bita, i, s[i] != 'A', n);
    if (s[i] == 'A')
      pa[i] = pa[i - 1] + 1;
    else
      pa[i] = 0;
  }
  for (int i = 1; i <= m; i++) {
    update(bitb, i, t[i] != 'A', m);
    if (t[i] == 'A')
      pb[i] = pb[i - 1] + 1;
    else
      pb[i] = 0;
  }
}
int main() {
  scanf("%s%s", s + 1, t + 1);
  init();
  int T;
  scanf("%d", &T);
  int cnt = 0;
  while (T--) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    int p = sum(bita, b) - sum(bita, a - 1);
    int q = sum(bitb, d) - sum(bitb, c - 1);
    int k1 = min(pa[b], b - a + 1);
    int k2 = min(pb[d], d - c + 1);
    if (p < q) {
      if ((q - p) & 1)
        ans[cnt] = 0;
      else {
        if (p == 0) {
          ans[cnt] = (k1 > k2);
        } else {
          if (k1 >= k2)
            ans[cnt] = 1;
          else
            ans[cnt] = 0;
        }
      }
    } else if (p > q)
      ans[cnt] = 0;
    else {
      if (k1 >= k2 && (k1 - k2) % 3 == 0)
        ans[cnt] = 1;
      else
        ans[cnt] = 0;
    }
    cnt++;
  }
  for (int i = 0; i < cnt; i++) printf("%d", ans[i]);
  return 0;
}
