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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const int INF = (int)(1e9 + 1e5);
const long long LINF = (long long)(1e18);
const long long MX = 100500;
const double EPS = 1e-9;
const int MOD = 1000000007;
const int ROOT = 700;
const int PRM_ROOT = 5;
const int LOG = 30;
const int MAXN = 100005;
struct query {
  int l1, r1, l2, r2;
};
int m;
char s[MAXN], t[MAXN];
query qu[MAXN];
int ps[MAXN], pt[MAXN];
string ans;
void init() {
  scanf("%s", &s[1]);
  scanf("%s", &t[1]);
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d%d%d", &qu[i].l1, &qu[i].r1, &qu[i].l2, &qu[i].r2);
  }
}
void build(char s[], int p[]) {
  for (int i = 1; s[i]; ++i) {
    p[i] = p[i - 1];
    if (s[i] == 'B' || s[i] == 'C') {
      ++p[i];
    }
  }
}
int get_conson(int p[], int l, int r) { return p[r] - p[l - 1]; }
int get_a(int p[], int lft, int rht) {
  int l = lft, r = rht, ans = rht + 1;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (get_conson(p, mid, r) == 0) {
      ans = mid;
      r = mid - 1;
    } else {
      l = mid + 1;
    }
  }
  return rht - ans + 1;
}
void solve() {
  build(s, ps);
  build(t, pt);
  ans.resize(m + 1);
  for (int i = 1; i <= m; ++i) {
    int cs = (get_conson(ps, qu[i].l1, qu[i].r1));
    int ct = (get_conson(pt, qu[i].l2, qu[i].r2));
    int as = (get_a(ps, qu[i].l1, qu[i].r1));
    int at = (get_a(pt, qu[i].l2, qu[i].r2));
    if ((cs & 1) != (ct & 1) || cs > ct) {
      ans[i] = '0';
    } else {
      if (cs) {
        ans[i] =
            (as % 3 == at % 3 && as >= at || cs < ct && as > at) ? '1' : '0';
      } else {
        ans[i] =
            (ct && at < as || !ct && at % 3 == as % 3 && as >= at) ? '1' : '0';
      }
    }
  }
  ans.erase(ans.begin());
}
void print_answer() { printf("%s\n", ans.c_str()); }
void run() {
  int t = 1;
  for (int i = 0; i < t; ++i) {
    init();
    solve();
    print_answer();
  }
}
int main() {
  srand(90);
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  run();
  return 0;
}
