#pragma GCC optimize ("O3")
#include <iostream>
#include <iomanip>
#include <istream>
#include <ostream>
#include <sstream>
#include <iterator>
#include <vector>
#include <algorithm>
#include <queue>
#include <deque>
#include <list>
#include <stack>
#include <map>
#include <unordered_map>
#include <set>
#include <bitset>
#include <utility>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <ctime>
#include <cctype>
#include <cstdlib>
#define IINF 10e8
#define INF 1<<30
#define MOD 1000000007
#define mod 1000000007
#define REP(i, a, n) for (ll i = a; i < (ll)(n); i++)
#define REPE(i, a, n) for (ll i = a; i <= (ll)(n); i++)
#define Endl endl
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define eb emplace_back
#define mmax(x,y)(x>y?x:y)
#define mmin(x,y)(x<y?x:y)
#define chmax(x,y) x=mmax(x,y)
#define chmin(x,y) x=mmin(x,y)
#define all(x) (x).begin(),(x).end()
#define siz(x) (ll)(x).size()
#define PI acos(-1.0)
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;
typedef long double ld;
typedef pair<int,int>Pin;
typedef pair<ll,ll>Pll;
template<class T> using V=vector<T>;
long long GCD(long long a, long long b) {return b?GCD(b,a%b):a;}
long long LCM(long long a, long long b) {return a/GCD(a,b)*b;}
int dx[4]={-1,0,1,0};
int dy[4]={0,-1,0,1};
int ddx[8]={-1,0,1,0,1,1,-1,-1};
int ddy[8]={0,-1,0,1,1,-1,1,-1};
ll cmp(pair<ll,ll>a,pair<ll,ll> b){
        if(a.se!=b.se)
        return a.se<b.se;
        else
        return a.fi<b.fi;
}
//----------------------------------------------------------------------
int n,k;
int rank_[1000010];
int tmp[1000010];
bool compare_sa(int i,int j){
    if(rank_[i]!=rank_[j])return rank_[i]<rank_[j];
    else{
        int ri=i+k<=n?rank_[i+k]:-1;
        int rj=j+k<=n?rank_[j+k]:-1;
        return ri<rj;
    }
}
void construct_sa(string S,int *sa){
    n=S.length();
    for(int i=0;i<=n;i++){
        sa[i]=i;
        rank_[i]=i<n?S[i]:-1;
    }
    for(k=1;k<=n;k*=2){
        sort(sa,sa+n+1,compare_sa);

        tmp[sa[0]]=0;
        for(int i=1;i<=n;i++){
            tmp[sa[i]]=tmp[sa[i-1]]+(compare_sa(sa[i-1],sa[i])?1:0);
        }
        for(int i=0;i<=n;i++){
            rank_[i]=tmp[i];
        }
    }
}

bool contain(string S, int *sa, string T){
    int a=0,b=S.length();
    while(b-a>1){
        int c=(a+b)/2;
        if(S.compare(sa[c],T.length(),T)<0)a=c;
        else b=c;
    }
    return S.compare(sa[b],T.length(),T)==0;
}
//----------------------------------------------------------------------
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    //-------------------------------  
    //ll begin_time=clock();
    //-------------------------------
    string S;cin>>S;
    int *sa=new int[S.length()+1];
    construct_sa(S,sa);
    int cnt;cin>>cnt;
    while(cnt--){
        string T;cin>>T;
        cout<<contain(S,sa,T)<<endl;
    }
    
    //-------------------------------  
    //ll end_time=clock();cout<<"time="<<end_time-begin_time<<"ms"<<endl;
    //-------------------------------
    return 0;
}
//----------------------------------------------------------------------














































### Prompt

Develop a solution in cpp to the problem described below:

Determine whether a text T includes a pattern P. Your program should answer for given queries consisting of P_i.

Constraints

* 1 ≤ length of T ≤ 1000000
* 1 ≤ length of P_i ≤ 1000
* 1 ≤ Q ≤ 10000
* The input consists of alphabetical characters and digits

Input

In the first line, a text T is given. In the second line, an integer Q denoting the number of queries is given. In the following Q lines, the patterns P_i are given respectively.

Output

For each question, print 1 if the text includes P_i, or print 0 otherwise.

Example

Input

aabaaa
4
aa
ba
bb
xyz


Output

1
1
0
0

### Response

```cpp
#pragma GCC optimize ("O3")
#include <iostream>
#include <iomanip>
#include <istream>
#include <ostream>
#include <sstream>
#include <iterator>
#include <vector>
#include <algorithm>
#include <queue>
#include <deque>
#include <list>
#include <stack>
#include <map>
#include <unordered_map>
#include <set>
#include <bitset>
#include <utility>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <ctime>
#include <cctype>
#include <cstdlib>
#define IINF 10e8
#define INF 1<<30
#define MOD 1000000007
#define mod 1000000007
#define REP(i, a, n) for (ll i = a; i < (ll)(n); i++)
#define REPE(i, a, n) for (ll i = a; i <= (ll)(n); i++)
#define Endl endl
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define eb emplace_back
#define mmax(x,y)(x>y?x:y)
#define mmin(x,y)(x<y?x:y)
#define chmax(x,y) x=mmax(x,y)
#define chmin(x,y) x=mmin(x,y)
#define all(x) (x).begin(),(x).end()
#define siz(x) (ll)(x).size()
#define PI acos(-1.0)
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;
typedef long double ld;
typedef pair<int,int>Pin;
typedef pair<ll,ll>Pll;
template<class T> using V=vector<T>;
long long GCD(long long a, long long b) {return b?GCD(b,a%b):a;}
long long LCM(long long a, long long b) {return a/GCD(a,b)*b;}
int dx[4]={-1,0,1,0};
int dy[4]={0,-1,0,1};
int ddx[8]={-1,0,1,0,1,1,-1,-1};
int ddy[8]={0,-1,0,1,1,-1,1,-1};
ll cmp(pair<ll,ll>a,pair<ll,ll> b){
        if(a.se!=b.se)
        return a.se<b.se;
        else
        return a.fi<b.fi;
}
//----------------------------------------------------------------------
int n,k;
int rank_[1000010];
int tmp[1000010];
bool compare_sa(int i,int j){
    if(rank_[i]!=rank_[j])return rank_[i]<rank_[j];
    else{
        int ri=i+k<=n?rank_[i+k]:-1;
        int rj=j+k<=n?rank_[j+k]:-1;
        return ri<rj;
    }
}
void construct_sa(string S,int *sa){
    n=S.length();
    for(int i=0;i<=n;i++){
        sa[i]=i;
        rank_[i]=i<n?S[i]:-1;
    }
    for(k=1;k<=n;k*=2){
        sort(sa,sa+n+1,compare_sa);

        tmp[sa[0]]=0;
        for(int i=1;i<=n;i++){
            tmp[sa[i]]=tmp[sa[i-1]]+(compare_sa(sa[i-1],sa[i])?1:0);
        }
        for(int i=0;i<=n;i++){
            rank_[i]=tmp[i];
        }
    }
}

bool contain(string S, int *sa, string T){
    int a=0,b=S.length();
    while(b-a>1){
        int c=(a+b)/2;
        if(S.compare(sa[c],T.length(),T)<0)a=c;
        else b=c;
    }
    return S.compare(sa[b],T.length(),T)==0;
}
//----------------------------------------------------------------------
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    //-------------------------------  
    //ll begin_time=clock();
    //-------------------------------
    string S;cin>>S;
    int *sa=new int[S.length()+1];
    construct_sa(S,sa);
    int cnt;cin>>cnt;
    while(cnt--){
        string T;cin>>T;
        cout<<contain(S,sa,T)<<endl;
    }
    
    //-------------------------------  
    //ll end_time=clock();cout<<"time="<<end_time-begin_time<<"ms"<<endl;
    //-------------------------------
    return 0;
}
//----------------------------------------------------------------------














































```