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 double eps = 1E-9;
const double Exp = 2.7182818284590452354;
const int Max_Bit = 63;
const int INF = 1000000007;
const long long LINF = 1000000000000000007ll;
const int MOD = 1000000007;
const int NMAX = 1e+5 + 5;
const int MMAX = 1e+7 + 5;
const int base = 1e+9;
string a, b;
int q, asum[NMAX], bsum[NMAX], acnt[NMAX], bcnt[NMAX];
void precalc() {
  for (int i = 0; i < a.length(); i++) {
    asum[i + 1] = asum[i];
    acnt[i + 1] = acnt[i];
    if (a[i] != 'A') {
      asum[i + 1]++;
      acnt[i + 1] = 0;
    } else {
      acnt[i + 1]++;
    }
  }
  for (int i = 0; i < b.length(); i++) {
    bsum[i + 1] = bsum[i];
    bcnt[i + 1] = bcnt[i];
    if (b[i] != 'A') {
      bsum[i + 1]++;
      bcnt[i + 1] = 0;
    } else {
      bcnt[i + 1]++;
    }
  }
}
int main() {
  cin >> a >> b;
  precalc();
  cin >> q;
  while (q--) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    int acntb = asum[b] - asum[a - 1];
    int bcntb = bsum[d] - bsum[c - 1];
    int acnta = min(b - a + 1, acnt[b]);
    int bcnta = min(d - c + 1, bcnt[d]);
    if (acntb == 0 && bcntb > 0) acnta = max(acnta - 1, 0);
    if (acntb > bcntb || (bcntb - acntb) % 2 != 0 || acnta < bcnta ||
        ((acnta - bcnta) % 3 != 0 && acntb >= bcntb))
      cout << 0;
    else
      cout << 1;
  }
  return 0;
}
