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;
string s, t;
int n;
int a[100010], b[100010], la[100010];
int at[100010], bt[100010], lat[100010];
int main() {
  cin >> s >> t;
  cin >> n;
  int ss = s.length(), tt = t.length();
  int cnt = 0;
  for (int i = 0; i < ss; i++) {
    if (s[i] == 'A') {
      a[i]++;
      cnt++;
    } else {
      b[i]++;
      cnt = 0;
    }
    la[i] = cnt;
  }
  cnt = 0;
  for (int i = 0; i < tt; i++) {
    if (t[i] == 'A') {
      at[i]++;
      cnt++;
    } else {
      bt[i]++;
      cnt = 0;
    }
    lat[i] = cnt;
  }
  for (int i = 1; i < ss; i++) a[i] += a[i - 1];
  for (int i = 1; i < ss; i++) b[i] += b[i - 1];
  for (int i = 1; i < tt; i++) at[i] += at[i - 1];
  for (int i = 1; i < tt; i++) bt[i] += bt[i - 1];
  int xs, ys, xt, yt;
  int sa, sb, ta, tb, sl, tl;
  for (int i = 0; i < n; i++) {
    cin >> xs >> ys >> xt >> yt;
    xs--;
    ys--;
    xt--;
    yt--;
    if (xs == 0) {
      sa = a[ys];
      sb = b[ys];
    } else {
      sa = a[ys] - a[xs - 1];
      sb = b[ys] - b[xs - 1];
    }
    if (xt == 0) {
      ta = at[yt];
      tb = bt[yt];
    } else {
      ta = at[yt] - at[xt - 1];
      tb = bt[yt] - bt[xt - 1];
    }
    if (sb == 0 && tb == 0) {
      if (sa < ta) {
        cout << 0;
        continue;
      }
      if ((sa - ta) % 3 == 0)
        cout << 1;
      else
        cout << 0;
      continue;
    }
    if (sb > tb) {
      cout << 0;
      continue;
    }
    if ((tb - sb) % 2 == 0) {
      sl = la[ys];
      tl = lat[yt];
      if (sl > ys - xs + 1) sl = ys - xs + 1;
      if (tl > yt - xt + 1) tl = yt - xt + 1;
      if (sl < tl) {
        cout << 0;
        continue;
      }
      if (sl == tl) {
        if (sl < ys - xs + 1 && tl < yt - xt + 1)
          cout << 1;
        else
          cout << 0;
        continue;
      }
      if (sb < tb) {
        cout << 1;
        continue;
      }
      if ((sl - tl) % 3 == 0)
        cout << 1;
      else
        cout << 0;
    } else
      cout << 0;
  }
  cout << endl;
  return 0;
}
