time limit per test : 2 seconds
memory limit per test : 256 megabytes

Alice has a string consisting of characters $'A'$′A′, $'B'$′B′ and $'C'$′C′. Bob can use the following transitions on any substring of our string in any order any number of times:

A -> BC
B -> AC
C -> AB
AAA -> 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 $S (1 ≤ |S| ≤ 10^5)$S(1 ≤ ∣S∣ ≤ 105). The second line contains a string $T (1 ≤ |T| ≤ 10^5)$T(1 ≤ ∣T∣ ≤ 105), each of these strings consists only of uppercase English letters $'A'$′A′, $'B'$′B′ and $'C'$′C′.

The third line contains the number of queries $Q (1 ≤ Q ≤ 10^5)$Q(1 ≤ Q ≤ 105).

The following $Q$Q lines describe queries. The $i$i-th of these lines contains four space separated integers $a_i, b_i, c_i, d_i$ai​,bi​,ci​,di​. These represent the $i$i-th query: is it possible to create $T[ci..di]$T[ci..di] from $S[ai..bi]$S[ai..bi] by applying the above transitions finite amount of times?

Here, $U[x..y]$U[x..y] is a substring of $U$U that begins at index $x$x (indexed from $1$1) and ends at index $y$y. In particular, $U[1..|U|]$U[1..∣U∣] is the whole string $U$U.

It is guaranteed that $1 ≤ a ≤ b ≤ |S|$1 ≤ a ≤ b ≤ ∣S∣ and $1 ≤ c ≤ d ≤ |T|$1 ≤ c ≤ d ≤ ∣T∣.

Output

Print a string of $Q$Q characters, where the $i$i-th character is $'1'$′1′ if the answer to the $i$i-th query is positive, and $'0'$′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 .

The third query asks for changing $AAB$AAB to $A$A — but in this case we are not able to get rid of the character $'B'$′B′.

题意：
给定字符变换规则：

A -> BC
B -> AC
C -> AB
AAA -> 空

给定两个长度为10万以内的字符串$S$S和$T$T，只有ABC三种字符。
一共有$q(q&lt;=10^5)$q(q<=105)组询问，每次询问$a,b,c,d$a,b,c,d
问字符串子串$S[a,b]$S[a,b]是否能通过若干次字符变换变成$T[c,d]$T[c,d]

题解：
首先我们可以知道

 B  <-> C

那么我们可以先将$A$A看作$1$1，$B$B/$C$C看作$0$0，然后将S串和T串都转换成01串。
我们发现，无论每个$0$0前面有几个$1$1，这些$1$1都是可以被消除的；而且每个$1$1可以变成$2*k（k&gt;=1）$2∗k（k>=1）个$0$0。
我们现在只需要考虑两个子串内的$0$0的个数和后缀的$1$1的个数即可，对于$0$0的个数差值为奇数的情况或者S串中0的个数大于T串中0的个数的情况，这些是不能转换的，对于结尾的连续的1的长度大于第二个串的情况，如果差值是3的倍数，那么刚好可以转换，否则的话需要判断0的差值是否大于等于2且为偶数。注意特判一下$S[a,b]$S[a,b]是全$1$1串的情况。

#include<bits/stdc++.h>
#define LiangJiaJun main
#define ll long long
using namespace std;
char s[2][100004],ans[100004];
int sum[2][100004],l[2];
int calc(int pt,int l,int r){
    return sum[pt][r]-sum[pt][l-1];
}

int getcon(int pt,int a,int b){
    int l=a,r=b,mid,ans=b+1;
    while(l<=r){
        mid=(l+r)>>1;
        if(calc(pt,mid,b)==b-mid+1){
            ans=mid;
            r=mid-1;
        }
        else l=mid+1;
    }
    return b-ans+1;
}

bool check(int a,int b,int c,int d){
     int g01=getcon(0,a,b),g11=getcon(1,c,d);
     int g00=b-a+1-calc(0,a,b),g10=d-c+1-calc(1,c,d);
     if(g01<g11||g00>g10)return 0;
     int delta=g01-g11;
     int gp=(delta%3>0);
     int sub=g10-g00;
     if(sub<(gp<<1))return 0;
     if(sub&1)return 0;
     if(g01==b-a+1&&sub>0&&g01==g11)return 0;
     return 1;
}
int LiangJiaJun(){
    scanf("%s",s[0]+1);
    scanf("%s",s[1]+1);
    l[0]=strlen(s[0]+1);
    l[1]=strlen(s[1]+1);
    sum[0][0]=0;
    sum[1][0]=0;
    for(int i=0;i<=1;i++){
        for(int j=1;j<=l[i];j++){
            sum[i][j]=sum[i][j-1]+(s[i][j]=='A');
        }
    }
    int q,len=0;scanf("%d",&q);
    for(int i=1;i<=q;i++){
        int a,b,c,d;
        scanf("%d%d%d%d",&a,&b,&c,&d);
        ans[i]=check(a,b,c,d)?'1':'0';
    }
    ans[q+1]='\n';
    puts(ans+1);
    return 0;
}