Problem:
Score : 600 points

Problem Statement

Let us consider the following operations on a string consisting of A and B:

Select a character in a string. If it is A, replace it with BB. If it is B, replace with AA.

Select a substring that is equal to either AAA or BBB, and delete it from the string.

For example, if the first operation is performed on ABA and the first character is selected, the string becomes BBBA.
If the second operation is performed on BBBAAAA and the fourth through sixth characters are selected, the string becomes BBBA.

These operations can be performed any number of times, in any order.

You are given two string S and T, and q queries a_i, b_i, c_i, d_i.
For each query, determine whether S_{a_i} S_{{a_i}+1} ... S_{b_i}, a substring of S, can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, a substring of T.

Constraints

1 \leq |S|, |T| \leq 10^5

S and T consist of letters A and B.

1 \leq q \leq 10^5

1 \leq a_i \leq b_i \leq |S|

1 \leq c_i \leq d_i \leq |T|

Input

Input is given from Standard Input in the following format:

S
T
q
a_1 b_1 c_1 d_1
...
a_q b_q c_q d_q

Output

Print q lines. The i-th line should contain the response to the i-th query. If S_{a_i} S_{{a_i}+1} ... S_{b_i} can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, print YES. Otherwise, print NO.

Sample Input 1

BBBAAAABA
BBBBA
4
7 9 2 5
7 9 1 4
1 7 2 5
1 7 2 4

Sample Output 1

YES
NO
YES
NO

The first query asks whether the string ABA can be made into BBBA.
As explained in the problem statement, it can be done by the first operation.

The second query asks whether ABA can be made into BBBB, and the fourth query asks whether BBBAAAA can be made into BBB.
Neither is possible.

The third query asks whether the string BBBAAAA can be made into BBBA.
As explained in the problem statement, it can be done by the second operation.

Sample Input 2

AAAAABBBBAAABBBBAAAA
BBBBAAABBBBBBAAAAABB
10
2 15 2 13
2 13 6 16
1 13 2 20
4 20 3 20
1 18 9 19
2 14 1 11
3 20 3 15
6 16 1 17
4 18 8 20
7 20 3 14

Sample Output 2

YES
YES
YES
YES
YES
YES
NO
NO
NO
NO

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <algorithm>
#include <queue>
#include <deque>
#include <stack>
#include <iomanip>
#include <cmath>


using namespace std;


typedef long long ll;
typedef long double ld;


const ll MOD_CONST = 1000000007ll;


ll modSum(ll a, ll b, ll MOD = MOD_CONST) {
  return ((a % MOD) + (b % MOD)) % MOD;
}


ll modSubtract(ll a, ll b, ll MOD = MOD_CONST) {
  return (((a % MOD) - (b % MOD)) + MOD + MOD) % MOD;
}


ll modProd(ll a, ll b, ll MOD = MOD_CONST) {
  return ((a % MOD) * (b % MOD)) % MOD;
}


ll getPowMod(ll x, ll e, ll MOD = MOD_CONST) {
  if (e == 0) return 1;
  if (e % 2 == 0) {
    ll tmp = getPowMod(x, e/2, MOD);
    return modProd(tmp, tmp);
  } else {
    ll tmp = getPowMod(x, e-1, MOD);
    return modProd(tmp, x);
  }
}


ll getPow(ll x, ll e) {
  if (e == 0) return 1;
  if (e % 2 == 0) {
    ll tmp = getPow(x, e/2);
    return tmp * tmp;
  } else {
    ll tmp = getPow(x, e-1);
    return tmp * x;
  }
}


ll getInverse(ll x, ll MOD = MOD_CONST) {
  return getPowMod(x, MOD-2, MOD);
}


bool isEven(ll x) {
  ll tmp = ((x % 2) + 2) % 2;
  return tmp == 0;
}


ll getSumOfDigitsInBase(ll n, ll b) {
  ll ret = 0;
  while (n > 0) {
    ret += n % b;
    n /= b;
  }
  return ret;
}


vector<int> getKMP(string &s) {
  int len = (int)s.size();
  vector<int> ret (len, 0);
  for (int i = 1 ; i < len ; i++) {
    int at = ret[i-1];
    while (at > 0 && s[i] != s[at]) {
      at = ret[at-1];
    }
    if (s[i] == s[at]) {
      at++;
    }
    ret[i] = at;
  }
  return ret;
}


string getSubstring(string &s, int from, int to) {
  int l = to-from+1;
  if (l <= 0) {
    return "";
  }
  return s.substr(from, l);
}

ll gcd(ll a, ll b, ll & x, ll & y) {
    if (a == 0) {
        x = 0;
        y = 1;
        return b;
    }
    ll x1, y1;
    ll d = gcd(b % a, a, x1, y1);
    x = y1 - (b / a) * x1;
    y = x1;
    return d;
}


pair<ll, ll> getIntersectingRange(ll a1, ll b1, ll a2, ll b2) {
  ll s = max(a1, a2);
  ll e = min(b1, b2);

  return {s, e};
}


bool isNonEmptyIntersection(ll a1, ll b1, ll a2, ll b2) {
  auto p = getIntersectingRange(a1, b1, a2, b2);
  return p.first <= p.second;
}


double getPointDistance(double x1, double y1, double x2, double y2) {
  double dx = x1-x2;
  double dy = y1-y2;
  double d = (dx * dx) + (dy * dy);
  return sqrt(d);
}


bool isPrime(ll x) {
  if (x == 2 || x == 3 || x == 5 || x == 7) return true;
  if (x < 10) return false;
  ll till = min((ll)sqrt(x) + 1, x-1);
  for (ll i = 2 ; i <= till ; i++) {
    if (x % i == 0) {
      return false;
    }
  }
  return true;
}


const int TREE_SIZE = 1;


ll segTree[TREE_SIZE], lazyTree[TREE_SIZE];


void updateRange(int node, int start, int end, int l, int r, int val)
{
    if(lazyTree[node] != 0)
    {
        // This node needs to be updated
        segTree[node] += (end - start + 1) * lazyTree[node];    // Update it
        if(start != end)
        {
            lazyTree[node*2] += lazyTree[node];                  // Mark child as lazyTree
            lazyTree[node*2+1] += lazyTree[node];                // Mark child as lazyTree
        }
        lazyTree[node] = 0;                                  // Reset it
    }
    if(start > end or start > r or end < l)              // Current segment is not within range [l, r]
        return;
    if(start >= l and end <= r)
    {
        // Segment is fully within range
        segTree[node] += (end - start + 1) * val;
        if(start != end)
        {
            // Not leaf node
            lazyTree[node*2] += val;
            lazyTree[node*2+1] += val;
        }
        return;
    }
    int mid = (start + end) / 2;
    updateRange(node*2, start, mid, l, r, val);        // Updating left child
    updateRange(node*2 + 1, mid + 1, end, l, r, val);   // Updating right child
    segTree[node] = segTree[node*2] + segTree[node*2+1];        // Updating root with max value
}


ll queryRange(int node, int start, int end, int l, int r)
{
    if(start > end or start > r or end < l)
        return 0;         // Out of range
    if(lazyTree[node] != 0)
    {
        // This node needs to be updated
        segTree[node] += (end - start + 1) * lazyTree[node];            // Update it
        if(start != end)
        {
            lazyTree[node*2] += lazyTree[node];         // Mark child as lazyTree
            lazyTree[node*2+1] += lazyTree[node];    // Mark child as lazyTree
        }
        lazyTree[node] = 0;                 // Reset it
    }
    if(start >= l and end <= r)             // Current segment is totally within range [l, r]
        return segTree[node];
    int mid = (start + end) / 2;
    int p1 = queryRange(node*2, start, mid, l, r);         // Query left child
    int p2 = queryRange(node*2 + 1, mid + 1, end, l, r); // Query right child
    return (p1 + p2);
}



















const int MAXN = 100001;


int Q;
string s, t;
int memo_sa[MAXN], memo_sb[MAXN], memo_ta[MAXN], memo_tb[MAXN];


int getClass(int from, int to, int * memo_a, int * memo_b) {

  int ca = memo_a[to] - memo_a[from-1];
  int cb = memo_b[to] - memo_b[from-1];

  return (ca + cb + cb) % 3;
}


void input() {
  cin >> s >> t;
  cin >> Q;
}


void preprocess() {
  for (int i = 1 ; i <= (int)s.size() ; i++) {
    memo_sa[i] = memo_sa[i-1] + (s[i-1] == 'A');
  }
  for (int i = 1 ; i <= (int)s.size() ; i++) {
    memo_sb[i] = memo_sb[i-1] + (s[i-1] == 'B');
  }
  for (int i = 1 ; i <= (int)t.size() ; i++) {
    memo_ta[i] = memo_ta[i-1] + (t[i-1] == 'A');
  }
  for (int i = 1 ; i <= (int)t.size() ; i++) {
    memo_tb[i] = memo_tb[i-1] + (t[i-1] == 'B');
  }
}


void solve() {
  int a, b, c, d;
  for (int i = 1 ; i <= Q ; i++) {
    cin >> a >> b >> c >> d;
    int c1 = getClass(a, b, memo_sa, memo_sb);
    int c2 = getClass(c, d, memo_ta, memo_tb);
    //cerr << "C1 = " << c1 << ", C2 = " << c2 << "\n";
    if (c1 == c2) {
      cout << "YES\n";
    } else {
      cout << "NO\n";
    }
  }
}


int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cerr.tie(0);

  int T;
//  cin >> T;
  T = 1;

  for (int i = 0 ; i < T ; i++) {
    input();
    preprocess();
    solve();
  }

  return 0;
}