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 inf = 1e9;
const long long INF = 1e18;
const long double PI = acos(-1);
const long double EPS = 1e-9;
int mod = 998244353;
const int MOD7 = 1000000007;
const int MOD9 = 1000000009;
mt19937 rnd(time(0));
const int MAXN = 1e5 + 228;
string s, t;
int stupid(int l1, int r1, int l2, int r2) {
  string sx = string(s.begin() + l1, s.begin() + r1 + 1);
  string tx = string(t.begin() + l2, t.begin() + r2 + 1);
  map<string, int> d;
  queue<string> q;
  q.push(sx);
  d[sx] = 1;
  while (q.size()) {
    string v = q.front();
    q.pop();
    if (v.size() > tx.size() * 2 + 2) {
      continue;
    }
    for (int i = 0; i < v.size(); ++i) {
      string tv = "";
      if (v[i] == 'A') {
        tv = v.substr(0, i);
        tv += "BB";
        for (int j = i + 1; j < v.size(); ++j) tv.push_back(v[j]);
        if (!d.count(tv)) {
          d[tv] = 1;
          q.push(tv);
        }
      } else if (v[i] == 'B') {
        tv = v.substr(0, i);
        tv += "AB";
        for (int j = i + 1; j < v.size(); ++j) tv.push_back(v[j]);
        if (!d.count(tv)) {
          d[tv] = 1;
          q.push(tv);
        }
      }
      if (i + 2 < v.size() && v.substr(i, 3) == "AAA") {
        tv = v.substr(0, i);
        for (int j = i + 3; j < v.size(); ++j) tv.push_back(v[j]);
        if (!d.count(tv)) {
          d[tv] = 1;
          q.push(tv);
        }
      }
    }
  }
  return d[tx];
  return 123213;
}
int prevb[MAXN];
int pref[MAXN];
int preft[MAXN];
int prevbt[MAXN];
void init() {
  for (int i = 0; i < s.size(); ++i)
    if (s[i] == 'C') s[i] = 'B';
  for (int i = 0; i < t.size(); ++i)
    if (t[i] == 'C') t[i] = 'B';
  for (int i = 0; i < s.size(); ++i) {
    if (s[i] == 'B') {
      prevb[i] = i;
    } else {
      if (i == 0)
        prevb[i] = -1;
      else
        prevb[i] = prevb[i - 1];
    }
  }
  for (int i = 0; i < t.size(); ++i) {
    if (t[i] == 'B')
      prevbt[i] = i;
    else {
      if (i == 0)
        prevbt[i] = -1;
      else
        prevbt[i] = prevbt[i - 1];
    }
  }
  for (int i = 0; i < s.size(); ++i) {
    pref[i] = (i == 0 ? 0 : pref[i - 1]) + (s[i] == 'B');
  }
  for (int i = 0; i < t.size(); ++i) {
    preft[i] = (i == 0 ? 0 : preft[i - 1]) + (t[i] == 'B');
  }
}
int smart(int l1, int r1, int l2, int r2) {
  int s1 = pref[r1] - (l1 == 0 ? 0 : pref[l1 - 1]);
  int s2 = preft[r2] - (l2 == 0 ? 0 : preft[l2 - 1]);
  if (s1 > s2) return 0;
  if (s1 % 2 != s2 % 2) return 0;
  int kekw = r1 - max(prevb[r1], l1 - 1);
  int kekt = r2 - max(prevbt[r2], l2 - 1);
  if (kekw < kekt) return 0;
  if (!s1 && s2) {
    if (kekw - 1 < kekt) return 0;
  }
  if (kekw % 3 != kekt % 3) {
    if (kekt % 3) {
      int go = kekw - kekt;
      go %= 3;
      if (s1 + 2 > s2) return 0;
    } else {
      int go = kekw - kekt;
      go %= 3;
      if (s1 + 2 > s2) return 0;
    }
  }
  return 1;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  srand(time(NULL));
  cin >> s >> t;
  int q;
  cin >> q;
  init();
  while (q--) {
    int l1, r1, l2, r2;
    cin >> l1 >> r1 >> l2 >> r2;
    l1--;
    r1--;
    l2--;
    r2--;
    cout << smart(l1, r1, l2, r2);
  }
}
