**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Trails (Hard)
time_limit_ms: 2000
memory_limit_mb: 256
Description: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.

Each day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.

How many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?

Give the answer modulo $$$10^9 + 7$$$.

Input: The first line contains the integers $$$m$$$ and $$$n$$$.

The second line contains $$$m$$$ integers, $$$s_1, \dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.

The third and last line contains $$$m$$$ integers, $$$l_1, \dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.

We have the following constraints:

$$$0 \le s_i, l_i \le 10^3$$$.

$$$1 \le m \le 10^5$$$.

$$$1 \le n \le 10^9$$$.

Output: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.

Examples:
input:
3 2
1 0 1
0 1 1
output:
18

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef long double ld;
const ll M = 1e9 + 7 ; // 998244353;

/*
**************************************************************************************************
**************************************************************************************************

                    ***        **************
                    ***        **************
                    ***        ***
                    ***        ***
                    ***        ***
                    *************************
                    *************************
                               ***        ***
                               ***        ***
                               ***        ***
                    **************        ***
                    **************        ***

**************************************************************************************************
**************************************************************************************************

*/

#define v(li) vector<ll>
#define vp(li) vector<pair<ll, ll>>
#define m(li) map<ll, ll>
#define s(li) set<ll>

#define f(i, n) for (ll i = 0; i < n; i++)
#define f1(i, n) for (ll i = 1; i < n+1; i++)
#define test(t) while (t--)
#define w(i, n) while (i < n)
#define prs(n) cout << n << " "
#define prn(n) cout << n << " "
#define newl cout << "\n"
#define pb(a) push_back(a);
#define pop pop_back()
#define in(x) insert(x)
#define fsort(v) sort(v.begin(), v.end())
#define rsort(v) sort(v.begin(), v.end(), greater<ll>())
#define rev(v) reverse(v.begin(), v.end())
#define max_ele(v) *max_element(v.begin(), v.end())
#define min_ele(v) *min_element(v.begin(), v.end())
#define prc(i) cout<<"Case #"<<i<<": ";

ll binexp(ll a, ll b)
{
    if (b == 0)
        return 1;
    ll res = binexp(a, b / 2);
    if (b % 2 == 0)
        return res * res;
    else
        return a * res * res;
}

ll biexm(ll a, ll b)
{
    if (b == 0)
        return 1;
    ll res = biexm(a, b / 2);
    if (b % 2 == 0)
        return ((res % M) * (res % M)) % M;
    else
        return ((((res % M) * (res % M)) % M) * (a % M)) % M;
}

ll hcf(ll a, ll b)
{
    return __gcd(a,b);
}
ll lcm(ll a,ll b){
    return (a*b)/(__gcd(a,b));
}

/*-------------------------------------------------------------------------------------------------------------------------------------------*/
// for ncr and factorial with modulo M

// const ll N1=3*1e5;
// ll fact[N1+1];
// ll inv[N1+1];

// ll ncr(ll n,ll r){
//     if(r>n) return 0;
//     ll temp=fact[n];
//     temp=(temp*inv[r])%M;
//     temp=(temp*inv[n-r])%M;
//     return temp;
// }

/*-------------------------------------------------------------------------------------------------------------------------------------------*/
/*  DSU  */

// const ll N2=2*1e5;
// ll par[N2+1];
// ll ran[N2+1];

// void make(ll n){
//     f1(i,n){
//         par[i]=i;
//         ran[i]=1;
//     }
// }

// ll findg(ll v){
//     if(par[v]==v) return v;
//     par[v]=findg(par[v]);
//     return par[v];
// }

// void unite(ll a,ll b){
//     a=findg(a);
//     b=findg(b);
//     if(a==b) return;
//     if(ran[a]<ran[b]) swap(a,b);
//     par[b]=a;
//     ran[a]=ran[a]+ran[b];
// }

/*-------------------------------------------------------------------------------------------------------------------------------------------*/
vector<vector<ll>> matMul(vector<vector<ll>> a,vector<vector<ll>> b){
    vector<vector<ll>> res(2,vector<ll>(2,0));
    for(int i=0;i<2;i++){
        for(int j=0;j<2;j++){
            for(int k=0;k<2;k++){
                res[i][j]+=(a[i][k]*b[k][j])%M;
                res[i][j]%=M;
            }
        }
    }
    return res;
}
vector<vector<ll>> matExm(vector<vector<ll>> a,ll n){
    vector<vector<ll>> res(2,vector<ll>(2,0));
    if(n==0){
        res[0][0]=1;res[1][1]=1;
        return res;
    }
    res=matExm(a,n/2);
    res=matMul(res,res);
    if(n%2==1) return matMul(a,res);
    else return res;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

/*-------------------------------------------------------------------------------------------------------------------------------------------*/
    /* Sieve */
    // ll N=1e6;
    // ll hpf[N+1];
    // f(i,N+1) hpf[i]=i;
    // for(ll i=2;i<N+1;i++){
    //     if(hpf[i]==i){
    //         for(ll j=i*i;j<N+1;j+=i){
    //            hpf[j]=i;
    //         }
    //     }
    // }

    // ll N=1e6;
    // bool isprime[N+1];
    // f1(i,N) isprime[i]=true;
    // isprime[1]=false;
    // for(ll i=2;i<N+1;i++){
    //     if(isprime[i]){
    //         for(ll j=i*i;j<N+1;j+=i) isprime[j]=false;
    //     }
    // }

    // vector<ll> primes;
    // for(ll i=2;i<N+1;i++) if(isprime[i]) primes.push_back(i);

    // fact[0]=1;
    // f1(i,N1) fact[i]=(fact[i-1]*i)%M;
    // f(i,N1+1) inv[i]=biexm(fact[i],M-2);

            /* Code Starts here */
            // Think of dp transitions very carefully
            // In greedy problems if it asks for some max or min then try to get some equations from given situation
            // if problem is related to swaps in permutations then think of cycles formed in permutation
            // In dijikstra algorithm always use visited array
            // radius of tree = ceil(diameter/2)

/* ---------------------------------------------------------------------------------------------------------------------------------------- */

    ll t=1;
    // cin>>t;
    test(t){
      ll m,n;
      cin>>m>>n;
      vector<ll> a(m),b(m);
      f(i,m) cin>>a[i];
      f(i,m) cin>>b[i];

        ll a2=0,b2=0,ab=0;
        ll x1=0,y1=0;
        f(i,m){
            ab+=a[i]*b[i];ab%=M;
            a2+=a[i]*a[i];a2%=M;
            b2+=b[i]*b[i];b2%=M;
            x1+=(a[i]+b[i]); x1%=M;
            y1+=(a[i]); y1%=M;
        }
        // cout<<ab<<' '<<a2<<' '<<b2<<'\n';
        ll const1 = (ab+a2)%M;
        ll const2 = (ab+b2)%M;

        vector<vector<ll>> v1={{const1,const2},{0,0}};
        vector<vector<ll>> v2={{a2,ab},{0,0}};
        vector<vector<ll>> v3=matExm(v1,n-1);
        vector<vector<ll>> v4=matExm(v2,n-1);
        // for(int i=0;i<2;i++){
        //     for(int j=0;j<2;j++) cout<<v4[i][j]<<' ';
        //     newl;
        // }
        ll x=(v3[0][0]*x1+v3[0][1]*y1)%M;
        ll y=(v4[0][0]*x1+v4[0][1]*y1)%M;
        ll ans=(a[0]*x+b[0]*y)%M;
        cout<<ans<<'\n';

    }

    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3 1
1 0 1
0 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Trails (Hard)
time_limit_ms: 2000
memory_limit_mb: 256
Description: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.

Each day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.

How many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?

Give the answer modulo $$$10^9 + 7$$$.

Input: The first line contains the integers $$$m$$$ and $$$n$$$.

The second line contains $$$m$$$ integers, $$$s_1, \dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.

The third and last line contains $$$m$$$ integers, $$$l_1, \dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.

We have the following constraints:

$$$0 \le s_i, l_i \le 10^3$$$.

$$$1 \le m \le 10^5$$$.

$$$1 \le n \le 10^9$$$.

Output: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.

Examples:
input:
3 2
1 0 1
0 1 1
output:
18

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;

const long long MOD = 1000000007;

vector<vector<int>> mult(vector<vector<int>> &a, vector<vector<int>> &b) {
    vector<vector<int>> mat(2, vector<int>(2));
    for (int i = 0; i < 2; i++) {
        for (int j = 0; j < 2; j++) {
            for (int k = 0; k < 2; k++) {
                mat[i][j] += a[i][k] * b[k][j];
                mat[i][j] %= MOD;
            }
        }
    }
    return mat;
}

vector<vector<int>> mat_exp(vector<vector<int>> &a, int e) {
    if (e == 1) {
        return a;
    }
    vector<vector<int>> half = mat_exp(a, e / 2);
    vector<vector<int>> res = mult(half, half);
    if (e % 2 == 1) {
        res = mult(res, a);
    }
    return res;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int m, n;
    cin >> m >> n;
    int s[m];
    int l[m];
    for (int i = 0; i < m; i++) {
        cin >> s[i];
    }

    long long ss = 0; long long sl = 0; long long ts = 0; long long tl = 0;
    long long tsum = 0; long long lsum = 0;
    for (int i = 0; i < m; i++) {
        cin >> l[i];
        ss += s[i] * s[i];
        sl += s[i] * l[i];
        tl += (s[i] + l[i]) * l[i];
        ts += (s[i] + l[i]) * s[i];
        tsum += s[i] + l[i];
        lsum += l[i];
    }
    ss %= MOD; sl %= MOD; tl %= MOD; ts %= MOD;

    vector<vector<int>> mat(2, vector<int>(2));
    mat[0][0] = ts; mat[0][1] = tl; mat[1][0] = ss; mat[1][1] = sl;

    mat_exp(mat, n);
    long long res = s[0] * mat[0][0] % MOD * tsum + s[0] * mat[0][1] % MOD * lsum
                  + s[1] * mat[1][0] % MOD * tsum + s[1] * mat[1][1] % MOD * lsum;
    res %= MOD;
    cout << res << "\n";
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
2 1
1 0
0 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Trails (Hard)
time_limit_ms: 2000
memory_limit_mb: 256
Description: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.

Each day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.

How many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?

Give the answer modulo $$$10^9 + 7$$$.

Input: The first line contains the integers $$$m$$$ and $$$n$$$.

The second line contains $$$m$$$ integers, $$$s_1, \dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.

The third and last line contains $$$m$$$ integers, $$$l_1, \dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.

We have the following constraints:

$$$0 \le s_i, l_i \le 10^3$$$.

$$$1 \le m \le 10^5$$$.

$$$1 \le n \le 10^9$$$.

Output: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.

Examples:
input:
3 2
1 0 1
0 1 1
output:
18

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef long double ld;
const ll M = 1e9 + 7 ; // 998244353;

/*
**************************************************************************************************
**************************************************************************************************

                    ***        **************
                    ***        **************
                    ***        ***
                    ***        ***
                    ***        ***
                    *************************
                    *************************
                               ***        ***
                               ***        ***
                               ***        ***
                    **************        ***
                    **************        ***

**************************************************************************************************
**************************************************************************************************

*/

#define v(li) vector<ll>
#define vp(li) vector<pair<ll, ll>>
#define m(li) map<ll, ll>
#define s(li) set<ll>

#define f(i, n) for (ll i = 0; i < n; i++)
#define f1(i, n) for (ll i = 1; i < n+1; i++)
#define test(t) while (t--)
#define w(i, n) while (i < n)
#define prs(n) cout << n << " "
#define prn(n) cout << n << " "
#define newl cout << "\n"
#define pb(a) push_back(a);
#define pop pop_back()
#define in(x) insert(x)
#define fsort(v) sort(v.begin(), v.end())
#define rsort(v) sort(v.begin(), v.end(), greater<ll>())
#define rev(v) reverse(v.begin(), v.end())
#define max_ele(v) *max_element(v.begin(), v.end())
#define min_ele(v) *min_element(v.begin(), v.end())
#define prc(i) cout<<"Case #"<<i<<": ";

ll binexp(ll a, ll b)
{
    if (b == 0)
        return 1;
    ll res = binexp(a, b / 2);
    if (b % 2 == 0)
        return res * res;
    else
        return a * res * res;
}

ll biexm(ll a, ll b)
{
    if (b == 0)
        return 1;
    ll res = biexm(a, b / 2);
    if (b % 2 == 0)
        return ((res % M) * (res % M)) % M;
    else
        return ((((res % M) * (res % M)) % M) * (a % M)) % M;
}

ll hcf(ll a, ll b)
{
    return __gcd(a,b);
}
ll lcm(ll a,ll b){
    return (a*b)/(__gcd(a,b));
}

/*-------------------------------------------------------------------------------------------------------------------------------------------*/
// for ncr and factorial with modulo M

// const ll N1=3*1e5;
// ll fact[N1+1];
// ll inv[N1+1];

// ll ncr(ll n,ll r){
//     if(r>n) return 0;
//     ll temp=fact[n];
//     temp=(temp*inv[r])%M;
//     temp=(temp*inv[n-r])%M;
//     return temp;
// }

/*-------------------------------------------------------------------------------------------------------------------------------------------*/
/*  DSU  */

// const ll N2=2*1e5;
// ll par[N2+1];
// ll ran[N2+1];

// void make(ll n){
//     f1(i,n){
//         par[i]=i;
//         ran[i]=1;
//     }
// }

// ll findg(ll v){
//     if(par[v]==v) return v;
//     par[v]=findg(par[v]);
//     return par[v];
// }

// void unite(ll a,ll b){
//     a=findg(a);
//     b=findg(b);
//     if(a==b) return;
//     if(ran[a]<ran[b]) swap(a,b);
//     par[b]=a;
//     ran[a]=ran[a]+ran[b];
// }

/*-------------------------------------------------------------------------------------------------------------------------------------------*/

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

/*-------------------------------------------------------------------------------------------------------------------------------------------*/
    /* Sieve */
    // ll N=1e6;
    // ll hpf[N+1];
    // f(i,N+1) hpf[i]=i;
    // for(ll i=2;i<N+1;i++){
    //     if(hpf[i]==i){
    //         for(ll j=i*i;j<N+1;j+=i){
    //            hpf[j]=i;
    //         }
    //     }
    // }

    // ll N=1e6;
    // bool isprime[N+1];
    // f1(i,N) isprime[i]=true;
    // isprime[1]=false;
    // for(ll i=2;i<N+1;i++){
    //     if(isprime[i]){
    //         for(ll j=i*i;j<N+1;j+=i) isprime[j]=false;
    //     }
    // }

    // vector<ll> primes;
    // for(ll i=2;i<N+1;i++) if(isprime[i]) primes.push_back(i);

    // fact[0]=1;
    // f1(i,N1) fact[i]=(fact[i-1]*i)%M;
    // f(i,N1+1) inv[i]=biexm(fact[i],M-2);

            /* Code Starts here */
            // Think of dp transitions very carefully
            // In greedy problems if it asks for some max or min then try to get some equations from given situation
            // if problem is related to swaps in permutations then think of cycles formed in permutation
            // In dijikstra algorithm always use visited array
            // radius of tree = ceil(diameter/2)

/* ---------------------------------------------------------------------------------------------------------------------------------------- */

    ll t=1;
    // cin>>t;
    test(t){
      ll m,n;
      cin>>m>>n;
      vector<ll> a(m),b(m);
      f(i,m) cin>>a[i];
      f(i,m) cin>>b[i];

        vector<ll> dpX(n+1,0),dpY(n+1,0);
        ll a2=0,b2=0,ab=0;
        f(i,m){
            ab+=a[i]*b[i];ab%=M;
            a2+=a[i]*a[i];a2%=M;
            b2+=b[i]*b[i];b2%=M;
            dpX[0]+=(a[i]+b[i]); dpX[0]%=M;
            dpY[0]+=(a[i]); dpY[0]%=M;
        }
        // cout<<ab<<' '<<a2<<' '<<b2<<'\n';
        ll const1 = (ab+a2)%M;
        ll const2 = (ab+b2)%M;

        f1(i,n){
            dpX[i]=((const1*dpX[i-1])%M + (const2*dpY[i-1])%M)%M;
            dpY[i]=((a2*dpX[i-1])%M + (ab*dpY[i-1])%M)%M;
        }
        ll ans= (a[0]*dpX[n-1]+b[0]*dpY[n-1])%M;
        cout<<ans<<'\n';

    }

    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
2 1
1 1
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Trails (Hard)
time_limit_ms: 2000
memory_limit_mb: 256
Description: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.

Each day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.

How many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?

Give the answer modulo $$$10^9 + 7$$$.

Input: The first line contains the integers $$$m$$$ and $$$n$$$.

The second line contains $$$m$$$ integers, $$$s_1, \dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.

The third and last line contains $$$m$$$ integers, $$$l_1, \dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.

We have the following constraints:

$$$0 \le s_i, l_i \le 10^3$$$.

$$$1 \le m \le 10^5$$$.

$$$1 \le n \le 10^9$$$.

Output: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.

Examples:
input:
3 2
1 0 1
0 1 1
output:
18

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define int long long

#define mod 1000000007

int s[200000],l[200000];
int t[200000];
int f[2000][2000];
int m,n;

vector<vector<int>> mmul(vector<vector<int>> a,vector<vector<int>> b, int n) {

	vector<vector<int>> res(n+5,vector<int> (n+5));

	for(int i=1; i<=n; i++) {
		for(int j=1; j<=n; j++) {
			for(int k=1; k<=n; k++) {
				res[i][j]=(res[i][j]+a[i][k]*b[k][j]%mod)%mod;
			}
		}
	}

	return res;
}

vector<vector<int>> mqpow(vector<vector<int>> a,int b, int n) {

	vector<vector<int>> res(n+5,vector<int> (n+5));
	for(int i=1; i<=n; i++) res[i][i]=1;

	while(b) {
		if(b&1) res=mmul(res,a,n);
		a=mmul(a,a,n);
		b/=2;
	}
	return res;
}

void E1() {
	cin>>n>>m;
	for(int i=1; i<=n; i++) cin>>s[i];
	for(int i=1; i<=n; i++) cin>>l[i],t[i]=s[i]+l[i];
	f[0][1]=1;
	for(int i=1; i<=m; i++) {
		for(int j=1; j<=n; j++) {
			for(int k=1; k<=n; k++) {
				f[i][k]+=f[i-1][j]*(t[j]*t[k]-l[j]*l[k])%mod;
			}
		}
	}
	int ans=0;
	for(int i=1; i<=n; i++) ans=(ans+f[m][i])%mod;
	cout<<ans<<"\n";
}

void E2() {
	cin>>n>>m;
	for(int i=1; i<=n; i++) cin>>s[i];
	for(int i=1; i<=n; i++) cin>>l[i];
	vector<vector<int>> f(n+5,vector<int> (n+5));

	for(int i=1; i<=n; i++) {
		for(int j=1; j<=n; j++) {
			f[i][j]=s[i]*s[j]+s[i]*l[j]+l[i]*s[j];
			f[i][j]%=mod;
		}
	}

	vector<vector<int>> ms=mqpow(f,m,n);
	vector<vector<int>> ans(n+5,vector<int> (n+5));

	ans[1][1]=1;

	ans=mmul(ans,ms,n);

	int res=0;
	for(int i=1; i<=n; i++) res=(res+ans[1][i])%mod;
	cout<<res<<'\n';

}

void E3() {
	cin>>n>>m;
	for(int i=1; i<=n; i++) cin>>s[i];
	for(int i=1; i<=n; i++) cin>>l[i],t[i]=s[i]+l[i];

	vector<vector<int>> B(n+5,vector<int> (3)), C(3,vector<int> (n+5));

	for(int i=1; i<=n; i++) {
		B[i][1]=t[i];
		B[i][2]=l[i];
		C[1][i]=t[i];
		C[2][i]=-l[i];
	}

	vector<vector<int>> ms(5, vector<int> (5));

	for(int i=1; i<=n; i++) {
		ms[1][1] = (ms[1][1] + C[1][i]*B[i][1] + mod) %mod;
		ms[1][2] = (ms[1][2] + C[1][i]*B[i][2] + mod) %mod;
		ms[2][1] = (ms[2][1] + C[2][i]*B[i][1] + mod) %mod;
		ms[2][2] = (ms[2][2] + C[2][i]*B[i][2] + mod) %mod;
	}

	ms=mqpow(ms,m-1,2);
	//cout<<ms[1][1]<<" "<<ms[1][2]<<"\n"<<ms[2][1]<<" "<<ms[2][2]<<"\n\n";

	vector<int> ans(n+5),tmp(n+5);
	vector<vector<int>> bb(n+5,vector<int>(5));
	ans[1]=1;

	for(int i=1; i<=n; i++) {
		bb[i][1] = (B[i][1]*ms[1][1] + B[i][2]*ms[2][1] +mod +mod)%mod;
		bb[i][2] = (B[i][1]*ms[1][2] + B[i][2]*ms[2][2] +mod +mod)%mod;
	}

	for(int i=1;i<=n;i++) {
		tmp[1]=(tmp[1]+ans[i]*bb[i][1] +mod)%mod;
		tmp[2]=(tmp[2]+ans[i]*bb[i][2] +mod)%mod;
	}

	int res=0;
	for(int i=1;i<=n;i++) {
		ans[i] = (tmp[1]*C[1][i]+tmp[2]*C[2][i] +mod +mod)%mod;
		res=(res+ans[i])%mod;
	}

	cout<<res<<'\n';

}

signed main() {

	E3();

	return 0;
}

// 4 2 6
// 1+1*1+1+

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
2 1
1 0
0 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Trails (Hard)
time_limit_ms: 2000
memory_limit_mb: 256
Description: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.

Each day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.

How many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?

Give the answer modulo $$$10^9 + 7$$$.

Input: The first line contains the integers $$$m$$$ and $$$n$$$.

The second line contains $$$m$$$ integers, $$$s_1, \dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.

The third and last line contains $$$m$$$ integers, $$$l_1, \dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.

We have the following constraints:

$$$0 \le s_i, l_i \le 10^3$$$.

$$$1 \le m \le 10^5$$$.

$$$1 \le n \le 10^9$$$.

Output: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.

Examples:
input:
3 2
1 0 1
0 1 1
output:
18

Note:

Buggy Code:
MOD = 10**9 + 7

def matrix_mult(A, B, mod):
    return [[sum(x * y for x, y in zip(row, col)) % mod for col in zip(*B)] for row in A]

def matrix_pow(mat, exp, mod):
    n = len(mat)
    res = [[1 if i == j else 0 for j in range(n)] for i in range(n)]
    base = mat
    while exp:
        if exp % 2:
            res = matrix_mult(res, base, mod)
        base = matrix_mult(base, base, mod)
        exp //= 2
    return res

def solve(m, n, s, l):
    # Create the transition matrix
    T = [[0] * m for _ in range(m)]
    for i in range(m):
        for j in range(m):
            if i == j:
                T[i][j] = (s[i] * s[j] + s[i] * l[j] + l[i] * s[j]) % MOD
            else:
                T[i][j] = (s[i] * s[j] + s[i] * l[j] + l[i] * s[j]) % MOD

    # Initial vector v, starting at cabin 1
    v = [[0] for _ in range(m)]
    v[0][0] = 1  # Harry starts at cabin 1

    # Compute T^n
    Tn = matrix_pow(T, n, MOD)

    # Compute T^n * v
    result_vector = matrix_mult(Tn, v, MOD)

    # Sum all ways to be at any cabin after n days
    result = sum(row[0] for row in result_vector) % MOD
    return result

# Read input
import sys
input = sys.stdin.read
data = input().split()
m = int(data[0])
n = int(data[1])
s = list(map(int, data[2:m+2]))
l = list(map(int, data[m+2:2*m+2]))

# Solve the problem
print(solve(m, n, s, l))

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases that can highlight issues in the logic. Specifically, we should consider the following scenarios:

1. **Cabins with no short trails**: This can lead to situations where the transition matrix might not handle the constraints correctly.
2. **Cabins with no long trails**: Similarly, this can cause issues if the transition matrix does not account for the constraints properly.
3. **Large number of days**: This can test the efficiency and correctness of the matrix exponentiation.
4. **Small number of cabins**: This can help in verifying the basic logic without the complexity of large inputs.

Given these considerations, a small-scale test input that is complete and valid can be:

```plaintext
2 3
1 0
1 1
```

### Explanation:
- **m = 2**: There are 2 cabins.
- **n = 3**: Harry walks for 3 days.
- **s = [1, 0]**: Cabin 1 has 1 short trail, and Cabin 2 has 0 short trails.
- **l = [1, 1]**: Cabin 1 has 1 long trail, and Cabin 2 has 1 long trail.

This input is designed to test the following:
- The transition matrix should correctly handle the case where Cabin 2 has no short trails.
- The matrix exponentiation should handle the case of 3 days, which is a small but non-trivial number of days.
- The output should be computed correctly, considering the constraints that at least one of the two trails must be short.

This test input should help in exposing any issues in the buggy code's logic for handling the constraints and transitions between cabins.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Trails (Hard)
time_limit_ms: 2000
memory_limit_mb: 256
Description: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.

Each day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.

How many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?

Give the answer modulo $$$10^9 + 7$$$.

Input: The first line contains the integers $$$m$$$ and $$$n$$$.

The second line contains $$$m$$$ integers, $$$s_1, \dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.

The third and last line contains $$$m$$$ integers, $$$l_1, \dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.

We have the following constraints:

$$$0 \le s_i, l_i \le 10^3$$$.

$$$1 \le m \le 10^5$$$.

$$$1 \le n \le 10^9$$$.

Output: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.

Examples:
input:
3 2
1 0 1
0 1 1
output:
18

Note:

Buggy Code:
import os, sys
from io import BytesIO, IOBase
from array import array
from itertools import accumulate
import bisect
import math
from collections import deque

# from functools import cache
# cache cf需要自己提交 pypy3.9!
from copy import deepcopy

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, 8192))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, 8192))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

input = lambda: sys.stdin.readline().strip()
ints = lambda: list(map(int, input().split()))
Int = lambda: int(input())

def queryInteractive(a, b, c):
    print('? {} {} {}'.format(a, b, c))
    sys.stdout.flush()
    return int(input())

def answerInteractive(x1, x2):
    print('! {} {}'.format(x1, x2))
    sys.stdout.flush()

inf = float('inf')
#if "PyPy" in sys.version:
#    import pypyjit; pypyjit.set_param('max_unroll_recursion=-1')
MOD = int(1e9 + 7)
m,n = ints()
s = ints()
l = ints()

def solve_easy():
    dp = [[0]*m for _ in range(n+1)]
    dp[0][0] = 1

    for i in range(1,n+1):
        for j in range(m):
            for k in range(m):
                dp[i][j] = (dp[i][j] + dp[i-1][k]*(s[k]*s[j] + s[k]*l[j] + l[k]*s[j])%MOD)%MOD
    # print(dp)
    print(sum(dp[n])%MOD)

def mat_mul(X,Y):
    n,m = len(X),len(Y[0])
    res = [[0 for j in range(m)] for i in range(n)]
    for i in range(n):
        for j in range(m):
            for k in range(len(Y)):
                res[i][j] += (X[i][k] * Y[k][j])%MOD
                res[i][j] %= MOD
    return res

I  = [[0]*m for _ in range(m)]
for i in range(m):
    I[i][i] = 1

def exp(x, i):
    ans = I

    tmp = x
    while i:
        if i & 1:
            ans = mat_mul(ans , tmp)
        i >>= 1
        tmp = mat_mul(tmp , tmp)
    return ans

def solve_med():
    P = [[0]*m for _ in range(m)]
    for i in range(m):
        for j in range(m):
            P[i][j] = (s[i]*s[j] + s[i]*l[j] + l[i]*s[j])%MOD
    ans = exp(P,n)
    t = 0
    for x in ans[0]:
        t = (t + x)%MOD
    print(t)

solve_med()

# import sys
# for line in sys.stdin:
#     a,b = map(int, line.split())
#     print(a+b)

# python A.py < i.txt > o.txt

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases and scenarios that might cause the algorithm to fail. Here are the steps to generate a challenging test input:

1. **Constraints and Edge Cases:**
   - The number of cabins `m` and the number of days `n` should be within the given constraints.
   - The number of short and long trails for each cabin should also be within the constraints.

2. **Identify Potential Bugs:**
   - The code uses matrix exponentiation to solve the problem, which can be computationally intensive.
   - The code might fail if the matrix operations are not handled correctly, especially for large values of `m` and `n`.

3. **Generate a Small-Scale Test Input:**
   - We need to create a test input that is small enough to be manageable but still challenging for the algorithm.
   - A good test input would be one where the number of cabins `m` is small, but the number of days `n` is large, to test the efficiency of the matrix exponentiation.

4. **Test Input:**
   - Let's choose `m = 3` and `n = 1000000000` (a large value to test the efficiency of the algorithm).
   - For the number of short and long trails, we can use values that are not zero to ensure that the algorithm has to handle non-trivial cases.

Here is the test input:

```plaintext
3 1000000000
1 2 3
4 5 6
```

This input will:
- Test the efficiency of the matrix exponentiation for a large value of `n`.
- Ensure that the algorithm correctly handles non-zero values for the number of short and long trails.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Trails (Hard)
time_limit_ms: 2000
memory_limit_mb: 256
Description: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.

Each day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.

How many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?

Give the answer modulo $$$10^9 + 7$$$.

Input: The first line contains the integers $$$m$$$ and $$$n$$$.

The second line contains $$$m$$$ integers, $$$s_1, \dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.

The third and last line contains $$$m$$$ integers, $$$l_1, \dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.

We have the following constraints:

$$$0 \le s_i, l_i \le 10^3$$$.

$$$1 \le m \le 10^5$$$.

$$$1 \le n \le 10^9$$$.

Output: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.

Examples:
input:
3 2
1 0 1
0 1 1
output:
18

Note:

Buggy Code:
import java.util.Scanner;

public class Test {
    static final int MOD = 1_000_000_007;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int m = sc.nextInt();
        int n = sc.nextInt();
        int[] s = new int[m];
        int[] l = new int[m];

        for (int i = 0; i < m; i++) {
            s[i] = sc.nextInt();
        }
        for (int i = 0; i < m; i++) {
            l[i] = sc.nextInt();
        }

        // Create the transition matrix T
        long[][] T = new long[m][m];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < m; j++) {
                long shortToShort = (long) s[i] * s[j] % MOD;
                long shortToLong = (long) s[i] * l[j] % MOD;
                long longToShort = (long) l[i] * s[j] % MOD;
                // Ensure at least one short trail in any round trip
                T[i][j] = (shortToShort + shortToLong + longToShort) % MOD;
            }
        }

        // Exponentiate the matrix T to the power n
        long[][] result = matrixExponentiation(T, n, m);

        // Calculate the sum of the first row of the resulting matrix
        long totalWays = 0;
        for (int j = 0; j < m; j++) {
            totalWays = (totalWays + result[0][j]) % MOD;
        }

        System.out.println(totalWays);
    }

    private static long[][] matrixExponentiation(long[][] matrix, int power, int size) {
        long[][] result = new long[size][size];
        // Initialize result as the identity matrix
        for (int i = 0; i < size; i++) {
            result[i][i] = 1;
        }

        long[][] base = matrix;

        while (power > 0) {
            if ((power & 1) == 1) {
                result = matrixMultiply(result, base, size);
            }
            base = matrixMultiply(base, base, size);
            power >>= 1;
        }

        return result;
    }

    private static long[][] matrixMultiply(long[][] A, long[][] B, int size) {
        long[][] product = new long[size][size];
        for (int i = 0; i < size; i++) {
            for (int j = 0; j < size; j++) {
                for (int k = 0; k < size; k++) {
                    product[i][j] = (product[i][j] + A[i][k] * B[k][j] % MOD) % MOD;
                }
            }
        }
        return product;
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases or scenarios that maximize the probability of failure. Here are the steps to think through this:

1. **Edge Case: No Short Trails in Any Cabin**:
   - If all cabins have no short trails, the code should not count any valid paths because at least one short trail is required in each round trip. However, the current code does not handle this case correctly.

2. **Edge Case: Very Large Number of Days**:
   - The code should handle a large number of days (up to \(10^9\)). This will test the matrix exponentiation part of the code.

3. **Edge Case: Single Cabin**:
   - If there is only one cabin, the only valid path is from cabin 1 to the lake and back to cabin 1. This should be a simple case to test the basic logic.

4. **Edge Case: All Cabins Have Only Long Trails**:
   - Similar to the first edge case, if all cabins have only long trails, no valid paths should exist. This will test the transition matrix logic.

5. **Edge Case: Mixed Trails**:
   - A mix of short and long trails can help test the transition matrix and the matrix exponentiation logic.

Given these considerations, let's create a small-scale test input that includes some of these edge cases:

```plaintext
3 2
0 0 0
1 1 1
```

This test input:
- Has 3 cabins.
- No cabin has any short trails.
- Each cabin has exactly one long trail.

This input should result in 0 valid paths because at least one short trail is required in each round trip. If the code does not handle this case correctly, it will expose the bug.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Trails (Hard)
time_limit_ms: 2000
memory_limit_mb: 256
Description: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.

Each day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.

How many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?

Give the answer modulo $$$10^9 + 7$$$.

Input: The first line contains the integers $$$m$$$ and $$$n$$$.

The second line contains $$$m$$$ integers, $$$s_1, \dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.

The third and last line contains $$$m$$$ integers, $$$l_1, \dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.

We have the following constraints:

$$$0 \le s_i, l_i \le 10^3$$$.

$$$1 \le m \le 10^5$$$.

$$$1 \le n \le 10^9$$$.

Output: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.

Examples:
input:
3 2
1 0 1
0 1 1
output:
18

Note:

Buggy Code:
import java.util.Scanner;

public class Test {
    static final int MOD = 1_000_000_007;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int m = sc.nextInt();
        int n = sc.nextInt();
        int[] s = new int[m];
        int[] l = new int[m];

        for (int i = 0; i < m; i++) {
            s[i] = sc.nextInt();
        }
        for (int i = 0; i < m; i++) {
            l[i] = sc.nextInt();
        }

        // Create the transition matrix T
        long[][] T = new long[m][m];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < m; j++) {
                long shortToShort = (long) s[i] * s[j] % MOD;
                long shortToLong = (long) s[i] * l[j] % MOD;
                long longToShort = (long) l[i] * s[j] % MOD;
                // Ensure at least one short trail in any round trip
                T[i][j] = (shortToShort + shortToLong + longToShort) % MOD;
            }
        }

        // Exponentiate the matrix T to the power n
        long[][] result = matrixExponentiation(T, n, m);

        // Calculate the sum of the first row of the resulting matrix
        long totalWays = 0;
        for (int j = 0; j < m; j++) {
            totalWays = (totalWays + result[0][j]) % MOD;
        }

        System.out.println(totalWays);
    }

    private static long[][] matrixExponentiation(long[][] matrix, int power, int size) {
        long[][] result = new long[size][size];
        // Initialize result as the identity matrix
        for (int i = 0; i < size; i++) {
            result[i][i] = 1;
        }

        while (power > 0) {
            if ((power & 1) == 1) {
                result = matrixMultiply(result, matrix, size);
            }
            matrix = matrixMultiply(matrix, matrix, size);
            power >>= 1;
        }

        return result;
    }

    private static long[][] matrixMultiply(long[][] A, long[][] B, int size) {
        long[][] product = new long[size][size];
        for (int i = 0; i < size; i++) {
            for (int j = 0; j < size; j++) {
                for (int k = 0; k < size; k++) {
                    product[i][j] = (product[i][j] + A[i][k] * B[k][j] % MOD) % MOD;
                }
            }
        }
        return product;
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
2 1
1 0
0 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Trails (Hard)
time_limit_ms: 2000
memory_limit_mb: 256
Description: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.

Each day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.

How many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?

Give the answer modulo $$$10^9 + 7$$$.

Input: The first line contains the integers $$$m$$$ and $$$n$$$.

The second line contains $$$m$$$ integers, $$$s_1, \dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.

The third and last line contains $$$m$$$ integers, $$$l_1, \dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.

We have the following constraints:

$$$0 \le s_i, l_i \le 10^3$$$.

$$$1 \le m \le 10^5$$$.

$$$1 \le n \le 10^9$$$.

Output: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.

Examples:
input:
3 2
1 0 1
0 1 1
output:
18

Note:

Buggy Code:
// LUOGU_RID: 170851215
#include <bits/stdc++.h>
const int mod = 1e9 + 7;
struct matrix {
	int n, m;
	std::vector<std::vector<int> > a;
	matrix(int n1) {
		n = m = n1;
		a.assign(n, std::vector<int>(n));
		for (int i = 0; i < n; ++i)
			a[i][i] = 1;
		return;
	}
	matrix(int n1, int m1): n(n1), m(m1), a(n, std::vector<int>(m)) {}
	std::vector<int> &operator[] (const int q) {
		return a[q];
	}
	matrix operator* (matrix q) const {
		matrix res(n, q.m);
		for (int i = 0; i < n; ++i)
			for (int j = 0; j < q.m; ++j)
				for (int k = 0; k < m; ++k)
					res[i][j] = (((long long)a[i][k] * q[k][j]) % mod + res[i][j]) % mod;
		return res;
	}
	matrix &operator*= (matrix &q) {
		return *this = *this * q;
	}
	matrix operator^ (const int q) const {
		matrix res(n), x(*this);
		for (int y(q); y; y >>= 1, x *= x)
			if (y & 1)
				res *= x;
		return res;
	}
	matrix &operator^= (const int q) {
		return *this = *this ^ q;
	}
};
int main() {
#ifdef ONLINE_JUDGE
	std::ios::sync_with_stdio(false);
	std::cin.tie(nullptr);
	std::cout.tie(nullptr);
#endif
	int n, m;
	std::cin >> m >> n;
	std::vector<int> s(m), l(m);
	for (int i = 0; i < m; ++i)
		std::cin >> s[i];
	for (int i = 0; i < m; ++i)
		std::cin >> l[i];
	matrix a(m, 2), b(2, m);
	for (int i = 0; i < m; ++i) {
		a[i][0] = s[i] + l[i];
		a[i][1] = -l[i];
		b[0][i] = s[i] + l[i];
		b[1][i] = l[i];
	}
	// a = (a * b) ^ n;
	a = a * ((b * a) ^ (n - 1)) * b;
	int res = 0;
	for (int i = 0; i < m; ++i)
		res = ((long long)res + a[0][i]) % mod;
	res = (res + mod) % mod;
	std::cout << res << '\n';
	return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. Here are the steps to think through the problem:

1. **Understand the Bug**: The bug in the code is likely related to the matrix operations and the way the transitions are being calculated. Specifically, the way the matrix `a` and `b` are defined and used in the matrix exponentiation might be incorrect.

2. **Edge Cases**:
   - **Single Cabin**: If there is only one cabin, the number of trails should be straightforward.
   - **No Short Trails**: If a cabin has no short trails, it should not be possible to complete the hike in a day if the other trail is also long.
   - **Large Number of Days**: If the number of days is large, the matrix exponentiation should handle it correctly.

3. **Small-Scale Test Input**: We need a small-scale test input that can highlight the issue. A good candidate is a scenario with a single cabin that has only long trails, which should result in zero possible combinations.

Given these considerations, let's generate a test input:

```plaintext
1 1
0
1
```

This test input represents a scenario with:
- 1 cabin.
- 1 day.
- 0 short trails.
- 1 long trail.

In this scenario, Harry cannot complete the hike in a day because he has no short trails to use. The expected output should be 0. However, the buggy code might not handle this edge case correctly and produce an incorrect result.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Trails (Hard)
time_limit_ms: 2000
memory_limit_mb: 256
Description: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.

Each day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.

How many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?

Give the answer modulo $$$10^9 + 7$$$.

Input: The first line contains the integers $$$m$$$ and $$$n$$$.

The second line contains $$$m$$$ integers, $$$s_1, \dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.

The third and last line contains $$$m$$$ integers, $$$l_1, \dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.

We have the following constraints:

$$$0 \le s_i, l_i \le 10^3$$$.

$$$1 \le m \le 10^5$$$.

$$$1 \le n \le 10^9$$$.

Output: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.

Examples:
input:
3 2
1 0 1
0 1 1
output:
18

Note:

Buggy Code:
#include <bits/stdc++.h>
#define ll long long
using namespace std;

const int mod = 1e9 + 7;
struct Matrix {
    vector<vector<ll>> data;
    int row() const { return data.size(); }
    int col() const { return data[0].size(); }
    vector<ll>& operator[](int i) { return data[i]; }
    const vector<ll>& operator[](int i) const { return data[i]; }
    Matrix(int r, int c) : data(r, vector<ll>(c)) { }

    static Matrix identity(int n) {
        Matrix a(n, n);
        for (int i = 0; i < n; ++i)
            a[i][i] = 1;
        return a;
    }

    Matrix operator*(const Matrix& b) const {
        int r = row(), c = b.col(), n = col();
        Matrix cMat(r, c);
        for (int i = 0; i < r; ++i)
            for (int j = 0; j < c; ++j)
                for (int k = 0; k < n; ++k)
     cMat[i][j] = (cMat[i][j] + (data[i][k] * b[k][j]+mod)%mod+mod)%mod;
        return cMat;
    }
    Matrix pow(int exp) const {
        Matrix base = *this;
        Matrix ans = Matrix::identity(row());
        while (exp > 0) {
            if (exp & 1)
                ans = ans * base;
            base = base * base;
            exp >>= 1;
        }
        return ans;
    }
};

int m, n;
vector<int> t, l;

signed main() {
     ios_base::sync_with_stdio(0);
     cin.tie(0);
     cout.tie(0);
    cin >> m >> n;
    t.resize(m);
    l.resize(m);

    for (int i = 0; i < m; ++i)
        cin >> t[i];
    for (int i = 0; i < m; ++i)
        cin >> l[i];
    for (int i = 0; i < m; ++i)
        t[i] += l[i];

    Matrix base(m, 1);
    base[0][0] = 1;
    Matrix B(m, 2);
    Matrix C(2, m);
    for (int i = 0; i < m; ++i) {
        B[i][0] = t[i];
        B[i][1] = l[i];
        C[0][i] = t[i];
        C[1][i] = -l[i];
    }
    Matrix kq = C * B;
    Matrix B_pow = kq.pow(n - 1);
    Matrix final_result = B * B_pow * C;

    ll res = 0;
    for (int i = 0; i < m; i++)
        res = (res + final_result[i][0]+mod) % mod;
    cout << res;
    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases or scenarios where the algorithm might fail. Here are some considerations:

1. **Small Number of Cabins and Days**: Since the problem constraints allow for a large number of cabins and days, a small-scale test input can still be challenging if it highlights specific edge cases.
2. **Short and Long Trails**: Ensure that the test input includes a mix of short and long trails to test the logic that enforces the constraint of at least one short trail per day.
3. **Initial Cabin**: The problem starts from cabin 1, so the test input should focus on this cabin.
4. **Modulo Operation**: Ensure the output is large enough to test the modulo operation.

Given these considerations, a small-scale test input that can expose the bug is:

```plaintext
2 3
1 0
1 1
```

### Explanation:
- **m = 2**: There are 2 cabins.
- **n = 3**: Harry walks for 3 days.
- **Short Trails**: Cabin 1 has 1 short trail, and Cabin 2 has 0 short trails.
- **Long Trails**: Cabin 1 has 1 long trail, and Cabin 2 has 1 long trail.

### Why This Input is Challenging:
1. **Cabin 1 to Lake and Back**: Cabin 1 has both a short and a long trail, so Harry can take both types of trails from Cabin 1 to the lake and back.
2. **Cabin 2 to Lake and Back**: Cabin 2 has only a long trail, so any combination involving Cabin 2 must include a short trail from another cabin.
3. **Multiple Days**: The 3 days allow for multiple combinations of trails, testing the logic that ensures at least one short trail per day.

This input should help expose any issues in the matrix multiplication and power calculation logic, particularly how the constraints are enforced over multiple days.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Trails (Hard)
time_limit_ms: 2000
memory_limit_mb: 256
Description: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.

Each day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.

How many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?

Give the answer modulo $$$10^9 + 7$$$.

Input: The first line contains the integers $$$m$$$ and $$$n$$$.

The second line contains $$$m$$$ integers, $$$s_1, \dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.

The third and last line contains $$$m$$$ integers, $$$l_1, \dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.

We have the following constraints:

$$$0 \le s_i, l_i \le 10^3$$$.

$$$1 \le m \le 10^5$$$.

$$$1 \le n \le 10^9$$$.

Output: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.

Examples:
input:
3 2
1 0 1
0 1 1
output:
18

Note:

Buggy Code:
import os, sys
from io import BytesIO, IOBase
from array import array
from itertools import accumulate
import bisect
import math
from collections import deque

# from functools import cache
# cache cf需要自己提交 pypy3.9!
from copy import deepcopy

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, 8192))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, 8192))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

input = lambda: sys.stdin.readline().strip()
ints = lambda: list(map(int, input().split()))
Int = lambda: int(input())

def queryInteractive(a, b, c):
    print('? {} {} {}'.format(a, b, c))
    sys.stdout.flush()
    return int(input())

def answerInteractive(x1, x2):
    print('! {} {}'.format(x1, x2))
    sys.stdout.flush()

inf = float('inf')
#if "PyPy" in sys.version:
#    import pypyjit; pypyjit.set_param('max_unroll_recursion=-1')
MOD = int(1e9 + 7)
m,n = ints()
s = ints()
l = ints()

def solve_easy():
    dp = [[0]*m for _ in range(n+1)]
    dp[0][0] = 1

    for i in range(1,n+1):
        for j in range(m):
            for k in range(m):
                dp[i][j] = (dp[i][j] + dp[i-1][k]*(s[k]*s[j] + s[k]*l[j] + l[k]*s[j])%MOD)%MOD
    # print(dp)
    print(sum(dp[n])%MOD)

def mat_mul(X,Y):
    n,m = len(X),len(Y[0])
    res = [[0 for j in range(m)] for i in range(n)]
    for i in range(n):
        for j in range(m):
            for k in range(len(Y)):
                res[i][j] += (X[i][k] * Y[k][j])%MOD
                res[i][j] %= MOD
    return res

I  = [[0]*2 for _ in range(2)]
for i in range(2):
    I[i][i] = 1

def exp(x, i):
    ans = I

    tmp = x
    while i:
        if i & 1:
            ans = mat_mul(ans , tmp)
        i >>= 1
        tmp = mat_mul(tmp , tmp)
    return ans

def solve_med():
    P = [[0]*m for _ in range(m)]
    for i in range(m):
        for j in range(m):
            P[i][j] = (s[i]*s[j] + s[i]*l[j] + l[i]*s[j])%MOD
    ans = exp(P,n)
    t = 0
    for x in ans[0]:
        t = (t + x)%MOD
    print(t)

def solve_hard():
    B = [[s[i]+l[i],l[i]] for i in range(m)]
    C = [[s[i]+l[i] for i in range(m)],[-l[i] for i in range(m)]]
    # print(mat_mul(C,B))
    t = exp(mat_mul(C,B),n-1)
    # print(t)
    ans = mat_mul(B,t)
    ans = mat_mul(ans,C)
    print(sum(ans[0])%MOD)
solve_hard()

# import sys
# for line in sys.stdin:
#     a,b = map(int, line.split())
#     print(a+b)

# python A.py < i.txt > o.txt

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes potential bugs in the provided code, we need to consider edge cases and scenarios that might cause issues. Here are some key points to consider:

1. **Edge Cases**:
   - **Single Cabin**: \( m = 1 \)
   - **Single Day**: \( n = 1 \)
   - **No Short Trails**: \( s_i = 0 \) for all \( i \)
   - **No Long Trails**: \( l_i = 0 \) for all \( i \)
   - **Maximum Trails**: \( s_i = 10^3 \) and \( l_i = 10^3 \) for all \( i \)

2. **Scenarios to Test**:
   - **Multiple Cabins with Different Trail Combinations**: Ensure that the code handles different combinations of short and long trails correctly.
   - **Large \( n \)**: Ensure that the code can handle large values of \( n \) efficiently.
   - **Zero Trails**: Test with cabins that have no trails to see how the code handles such cases.

Given the constraints and the need for a small-scale test input, let's create a test case that includes a mix of these scenarios:

- \( m = 3 \) (a small number of cabins)
- \( n = 10 \) (a moderate number of days)
- Cabins with a mix of short and long trails

Here is the test input:

```plaintext
3 10
1 0 1
0 1 1
```

This test case includes:
- Cabin 1 with 1 short trail and 0 long trails.
- Cabin 2 with 0 short trails and 1 long trail.
- Cabin 3 with 1 short trail and 1 long trail.

This input will test how the code handles a combination of cabins with and without short trails, and how it handles a moderate number of days. It should expose any issues with the logic for calculating the number of possible combinations of trails.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Trails (Hard)
time_limit_ms: 2000
memory_limit_mb: 256
Description: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.

Each day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.

How many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?

Give the answer modulo $$$10^9 + 7$$$.

Input: The first line contains the integers $$$m$$$ and $$$n$$$.

The second line contains $$$m$$$ integers, $$$s_1, \dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.

The third and last line contains $$$m$$$ integers, $$$l_1, \dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.

We have the following constraints:

$$$0 \le s_i, l_i \le 10^3$$$.

$$$1 \le m \le 10^5$$$.

$$$1 \le n \le 10^9$$$.

Output: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.

Examples:
input:
3 2
1 0 1
0 1 1
output:
18

Note:

Buggy Code:
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<string>
#include<cstring>
#include<algorithm>
#include<stack>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<bitset>
#define rep(i,a,b) for(int i=a;i<=(b);++i)
#define frep(i,a,b) for(int i=a;i<(b);++i)
#define drep(i,a,b) for(int i=a;i>=(b);--i)
#define ls tree[p].l
#define rs tree[p].r
#define mid ((l + r) >> 1)
typedef long long ll;
typedef unsigned long long ull;
#define Maxq priority_queue<int,vector<int>,greater<int> >
using namespace std;
int mylog(int a){
	int ans=0;
	if(a&0xffff0000){
		ans+=16;
		a>>=16;
	}
	if(a&0x0000ff00){
		ans+=8;
		a>>=8;
	}
	if(a&0x000000f0){
		ans+=4;
		a>>=4;
	}
	if(a&0x0000000c){
		ans+=2;
		a>>=2;
	}
	if(a&0x00000002){
		ans+=1;
		a>>=1;
	}
	return ans;
}
inline int read(){
	register int a=0,b=0;
	register char c;
	c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-')b=1;
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		a*=10;
		a+=c-'0';
		c=getchar();
	}
	return b?-a:a;
}
const int MAXM = 100010;
const int MOD = 1e9 + 7;
struct Matrix{
	int n, m;
	int **data;
	Matrix(int n, int m) : n(n), m(m){
		data = new int*[n + 1];
		rep(i, 0, n){
			data[i] = new int[m + 1];
		}
	}
	Matrix(const Matrix & other) : n(other.n), m(other.m){
		data = new int*[n + 1];
		rep(i, 0, n){
			data[i] = new int[m + 1];
		}
		rep(i, 1, n){
			rep(j, 1, m){
				data[i][j] = other.data[i][j];
			}
		}
	}
	~Matrix(){
		rep(i, 0, n){
			delete[] data[i];
		}
		delete[] data;
	}
	Matrix& operator= (const Matrix & other){
		if(&other == this) return *this;
		rep(i, 0, n){
			delete[] data[i];
		}
		delete[] data;
		n = other.n;
		m = other.m;
		data = new int*[n + 1];
		rep(i, 0, n){
			data[i] = new int[m + 1];
		}
		rep(i, 1, n){
			rep(j, 1, m){
				data[i][j] = other.data[i][j];
			}
		}
		return *this;
	}
	Matrix operator* (const Matrix &other){
		Matrix ans(n, other.m);
//		ans.n = n;
//		ans.m = other.m;
		rep(i, 1, ans.n){
			rep(j, 1, ans.m){
				ans.data[i][j] = 0;
				rep(k, 1, m){
					ans.data[i][j] = ((ll)ans.data[i][j] + (ll)data[i][k] * other.data[k][j] % MOD) % MOD;
					if(ans.data[i][j] < 0){
						cout << 1;
					}
				}
			}
		}
		return ans;
	}
};
Matrix ksm(Matrix a, int k){
	if(k == 1) return a;
	if(k & 1){
		Matrix tmp = ksm(a, k >> 1);
		return tmp * tmp * a;
	}
	else{
		Matrix tmp = ksm(a, k >> 1);
		return tmp * tmp;
	}
}
int s[MAXM], l[MAXM];
signed main(){
	int m = read(), n = read();
	rep(i, 1, m) s[i] = read();
	rep(i, 1, m) l[i] = read();
	Matrix A(m, 2), B(2, m);
	rep(i, 1, m){
		A.data[i][1] = s[i] + l[i];
		A.data[i][2] = l[i];
	}
	rep(i, 1, m){
		B.data[1][i] = s[i] + l[i];
		B.data[2][i] = MOD - l[i];
	}
	Matrix tmp = ksm(B * A, n - 1);
	Matrix ans = A * tmp * B;
	ll sum = 0;
	rep(i, 1, m) sum += ans.data[i][1];
	printf("%lld", sum % MOD);
    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
2 1
1 1
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Trails (Hard)
time_limit_ms: 2000
memory_limit_mb: 256
Description: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.

Each day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.

How many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?

Give the answer modulo $$$10^9 + 7$$$.

Input: The first line contains the integers $$$m$$$ and $$$n$$$.

The second line contains $$$m$$$ integers, $$$s_1, \dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.

The third and last line contains $$$m$$$ integers, $$$l_1, \dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.

We have the following constraints:

$$$0 \le s_i, l_i \le 10^3$$$.

$$$1 \le m \le 10^5$$$.

$$$1 \le n \le 10^9$$$.

Output: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.

Examples:
input:
3 2
1 0 1
0 1 1
output:
18

Note:

Buggy Code:
#include <bits/stdc++.h>
#define int long long
using namespace std;

const int MOD = 1e9 + 7;

inline int mod(int a) {
    return (a % MOD + MOD) % MOD;
}

struct Matrix {
    vector<vector<int>> data;

    int row() const { return data.size(); }
    int col() const { return data[0].size(); }

    vector<int>& operator[](int i) { return data[i]; }
    const vector<int>& operator[](int i) const { return data[i]; }

    Matrix(int r, int c) : data(r, vector<int>(c)) { }

    static Matrix identity(int n) {
        Matrix a(n, n);
        for (int i = 0; i < n; ++i)
            a[i][i] = 1;
        return a;
    }

    Matrix operator*(const Matrix& b) const {
        int r = row(), c = b.col(), n = col();
        Matrix cMat(r, c);
        for (int i = 0; i < r; ++i) {
            for (int j = 0; j < c; ++j) {
                cMat[i][j] = 0;
                for (int k = 0; k < n; ++k) {
                    cMat[i][j] = mod(cMat[i][j] + mod(data[i][k] * b[k][j]));
                }
            }
        }
        return cMat;
    }

    Matrix pow(int exp) const {
        Matrix base = *this;
        Matrix ans = Matrix::identity(row());
        while (exp > 0) {
            if (exp & 1)
                ans = ans * base;
            base = base * base;
            exp >>= 1;
        }
        return ans;
    }
};

int m, n;
vector<int> t, l;

signed main() {
     ios_base::sync_with_stdio(0);
     cin.tie(0);
     cout.tie(0);
    cin >> m >> n;
    t.resize(m);
    l.resize(m);

    for (int i = 0; i < m; ++i)
        cin >> t[i];
    for (int i = 0; i < m; ++i)
        cin >> l[i];
    for (int i = 0; i < m; ++i)
        t[i] += l[i];

    Matrix base(m, 1);
    base[0][0] = 1;
    Matrix B(m, 2);
    Matrix C(2, m);
    for (int i = 0; i < m; ++i) {
        B[i][0] = t[i];
        B[i][1] = l[i];
        C[0][i] = t[i];
        C[1][i] = -l[i];
    }
    Matrix kq = C * B;
    Matrix final_result = B * kq.pow(n - 1) * C;

    int res = 0;
    for (int i = 0; i < m; i++)
        res = (res + final_result[i][0]) % MOD;

    cout << res;
    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
2 1
1 0
0 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Trails (Hard)
time_limit_ms: 2000
memory_limit_mb: 256
Description: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.

Each day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.

How many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?

Give the answer modulo $$$10^9 + 7$$$.

Input: The first line contains the integers $$$m$$$ and $$$n$$$.

The second line contains $$$m$$$ integers, $$$s_1, \dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.

The third and last line contains $$$m$$$ integers, $$$l_1, \dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.

We have the following constraints:

$$$0 \le s_i, l_i \le 10^3$$$.

$$$1 \le m \le 10^5$$$.

$$$1 \le n \le 10^9$$$.

Output: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.

Examples:
input:
3 2
1 0 1
0 1 1
output:
18

Note:

Buggy Code:
#include <bits/stdc++.h>
#define int long long
using namespace std;

const int MOD = 1e9 + 7;

inline int mod(int a) {
    return (a % MOD + MOD) % MOD;
}

struct Matrix {
    vector<vector<int>> data;

    int row() const { return data.size(); }
    int col() const { return data[0].size(); }

    vector<int>& operator[](int i) { return data[i]; }
    const vector<int>& operator[](int i) const { return data[i]; }

    Matrix(int r, int c) : data(r, vector<int>(c)) { }

    static Matrix identity(int n) {
        Matrix a(n, n);
        for (int i = 0; i < n; ++i)
            a[i][i] = 1;
        return a;
    }

    Matrix operator*(const Matrix& b) const {
        int r = row(), c = b.col(), n = col();
        Matrix cMat(r, c);
        for (int i = 0; i < r; ++i) {
            for (int j = 0; j < c; ++j) {
                cMat[i][j] = 0;
                for (int k = 0; k < n; ++k) {
                    cMat[i][j] = mod(cMat[i][j] + mod(data[i][k] * b[k][j]));
                }
            }
        }
        return cMat;
    }

    Matrix pow(int exp) const {
        Matrix base = *this;
        Matrix ans = Matrix::identity(row());
        while (exp > 0) {
            if (exp & 1)
                ans = ans * base;
            base = base * base;
            exp >>= 1;
        }
        return ans;
    }
};

int m, n;
vector<int> t, l;

signed main() {
     ios_base::sync_with_stdio(0);
     cin.tie(0);
     cout.tie(0);
    cin >> m >> n;
    t.resize(m);
    l.resize(m);

    for (int i = 0; i < m; ++i)
        cin >> t[i];
    for (int i = 0; i < m; ++i)
        cin >> l[i];
    for (int i = 0; i < m; ++i)
        t[i] += l[i];

    Matrix base(m, 1);
    base[0][0] = 1;
    Matrix B(m, 2);
    Matrix C(2, m);
    for (int i = 0; i < m; ++i) {
        B[i][0] = t[i];
        B[i][1] = l[i];
        C[0][i] = t[i];
        C[1][i] = -l[i];
    }
    Matrix kq = C * B;
    Matrix B_pow = kq.pow(n - 1);
    Matrix final_result = B * B_pow * C;

    int res = 0;
    for (int i = 0; i < m; i++)
        res = (res + final_result[i][0]) % MOD;

    cout << res;
    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that could cause the algorithm to fail. Here are the steps to think through this:

1. **Edge Cases for Input Values:**
   - Minimum values for \( m \) and \( n \).
   - Maximum values for \( m \) and \( n \).
   - Cases where \( s_i \) and \( l_i \) are zero.
   - Cases where \( s_i \) and \( l_i \) are at their maximum values.

2. **Behavior of the Algorithm:**
   - The algorithm involves matrix multiplication and exponentiation.
   - The bug might be in how the matrix is constructed or how the exponentiation is handled.

3. **Matrix Construction:**
   - The matrix \( B \) and \( C \) are constructed in a way that might not correctly represent the problem constraints.
   - Specifically, the construction of \( C \) with negative values for \( l_i \) might be problematic.

4. **Exponentiation:**
   - The exponentiation of the matrix \( kq \) might not handle large values of \( n \) correctly.

Given these considerations, a small-scale test input that can expose the bug is:

- \( m = 2 \) and \( n = 2 \).
- \( s_1 = 1 \), \( s_2 = 0 \).
- \( l_1 = 0 \), \( l_2 = 1 \).

This input is small enough to be manageable and has a mix of short and long trails, which should challenge the matrix construction and exponentiation steps.

Here is the test input:

```plaintext
2 2
1 0
0 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Array Divisibility
time_limit_ms: 1000
memory_limit_mb: 256
Description: An array of integers $$$a_1,a_2,\cdots,a_n$$$ is beautiful subject to an integer $$$k$$$ if it satisfies the following:

- The sum of $$$a_{j}$$$ over all $$$j$$$ such that $$$j$$$ is a multiple of $$$k$$$ and $$$1 \le j \le n $$$, itself, is a multiple of $$$k$$$.
- More formally, if $$$\sum_{k | j} a_{j}$$$ is divisible by $$$k$$$ for all $$$1 \le j \le n$$$ then the array $$$a$$$ is beautiful subject to $$$k$$$. Here, the notation $$${k|j}$$$ means $$$k$$$ divides $$$j$$$, that is, $$$j$$$ is a multiple of $$$k$$$.

It can be shown that an answer always exists.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 100$$$). The description of the test cases follows.

The first and only line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$) — the size of the array.

Output: For each test case, print the required array as described in the problem statement.

Examples:
input:
3
3
6
7
output:
4 22 18
10 6 15 32 125 54
23 18 27 36 5 66 7

Note: In the second test case, when $$$n = 6$$$, for all integers $$$k$$$ such that $$$1 \le k \le 6$$$, let $$$S$$$ be the set of all indices of the array that are divisible by $$$k$$$.

- When $$$k = 1$$$, $$$S = \{1, 2, 3,4,5,6\}$$$ meaning $$$a_1+a_2+a_3+a_4+a_5+a_6=242$$$ must be divisible by $$$1$$$.
- When $$$k = 2$$$, $$$S = \{2,4,6\}$$$ meaning $$$a_2+a_4+a_6=92$$$ must be divisible by $$$2$$$.
- When $$$k = 3$$$, $$$S = \{3,6\}$$$ meaning $$$a_3+a_6=69$$$ must divisible by $$$3$$$.
- When $$$k = 4$$$, $$$S = \{4\}$$$ meaning $$$a_4=32$$$ must divisible by $$$4$$$.
- When $$$k = 5$$$, $$$S = \{5\}$$$ meaning $$$a_5=125$$$ must divisible by $$$5$$$.
- When $$$k = 6$$$, $$$S = \{6\}$$$ meaning $$$a_6=54$$$ must divisible by $$$6$$$.

Buggy Code:
#include <iostream>
using namespace std;

const int N = 1001;

void solve()
{
	int n;
	cin >> n;
	int a[N];
 	for (int i = 1; i <= n; i++) a[i] = i;
	for (int k = 2; k <= n; k++) {
		for (int c = 1; c * k <= n; c++) {
			int id = c * k;
			a[id] *= k;
		}
	}

	for (int i = 1; i <= n; i++) cout << a[i] << " ";
	cout << "\n";
}

int main()
{
	int T;
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin >> T;
	while (T--)
	{
		solve();
	}
	return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The buggy code multiplies elements at indices that are multiples of `k` by `k` for all `k` from 2 to `n`. This can lead to incorrect results, especially for small values of `n` where the multiplications can quickly grow very large and violate the divisibility conditions.

Let's consider the following test case:

1. **Small `n` with a few elements**: This will help us see the impact of the multiplications more clearly.
2. **Multiple test cases**: To ensure the code handles multiple test cases correctly.

Here is a small-scale test input that should expose the bug:

```plaintext
3
4
5
6
```

### Explanation:
- **Test Case 1 (`n = 4`)**:
  - The array should be `[1, 2, 3, 4]` initially.
  - For `k = 2`, the elements at indices 2 and 4 are multiplied by 2, resulting in `[1, 4, 3, 8]`.
  - For `k = 3`, the element at index 3 is multiplied by 3, resulting in `[1, 4, 9, 8]`.
  - For `k = 4`, the element at index 4 is multiplied by 4, resulting in `[1, 4, 9, 32]`.
  - The sums for `k = 1, 2, 3, 4` are `[46, 44, 9, 32]`, which are not all divisible by their respective `k`.

- **Test Case 2 (`n = 5`)**:
  - The array should be `[1, 2, 3, 4, 5]` initially.
  - For `k = 2`, the elements at indices 2, 4 are multiplied by 2, resulting in `[1, 4, 3, 8, 5]`.
  - For `k = 3`, the element at index 3 is multiplied by 3, resulting in `[1, 4, 9, 8, 5]`.
  - For `k = 4`, the element at index 4 is multiplied by 4, resulting in `[1, 4, 9, 32, 5]`.
  - For `k = 5`, the element at index 5 is multiplied by 5, resulting in `[1, 4, 9, 32, 25]`.
  - The sums for `k = 1, 2, 3, 4, 5` are `[71, 44, 9, 32, 25]`, which are not all divisible by their respective `k`.

- **Test Case 3 (`n = 6`)**:
  - The array should be `[1, 2, 3, 4, 5, 6]` initially.
  - For `k = 2`, the elements at indices 2, 4, 6 are multiplied by 2, resulting in `[1, 4, 3, 8, 5, 12]`.
  - For `k = 3`, the elements at indices 3, 6 are multiplied by 3, resulting in `[1, 4, 9, 8, 5, 36]`.
  - For `k = 4`, the element at index 4 is multiplied by 4, resulting in `[1, 4, 9, 32, 5, 36]`.
  - For `k = 5`, the element at index 5 is multiplied by 5, resulting in `[1, 4, 9, 32, 25, 36]`.
  - For `k = 6`, the element at index 6 is multiplied by 6, resulting in `[1, 4, 9, 32, 25, 216]`.
  - The sums for `k = 1, 2, 3, 4, 5, 6` are `[387, 256, 45, 32, 25, 216]`, which are not all divisible by their respective `k`.

This test input should help expose the bug in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Array Divisibility
time_limit_ms: 1000
memory_limit_mb: 256
Description: An array of integers $$$a_1,a_2,\cdots,a_n$$$ is beautiful subject to an integer $$$k$$$ if it satisfies the following:

- The sum of $$$a_{j}$$$ over all $$$j$$$ such that $$$j$$$ is a multiple of $$$k$$$ and $$$1 \le j \le n $$$, itself, is a multiple of $$$k$$$.
- More formally, if $$$\sum_{k | j} a_{j}$$$ is divisible by $$$k$$$ for all $$$1 \le j \le n$$$ then the array $$$a$$$ is beautiful subject to $$$k$$$. Here, the notation $$${k|j}$$$ means $$$k$$$ divides $$$j$$$, that is, $$$j$$$ is a multiple of $$$k$$$.

It can be shown that an answer always exists.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 100$$$). The description of the test cases follows.

The first and only line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$) — the size of the array.

Output: For each test case, print the required array as described in the problem statement.

Examples:
input:
3
3
6
7
output:
4 22 18
10 6 15 32 125 54
23 18 27 36 5 66 7

Note: In the second test case, when $$$n = 6$$$, for all integers $$$k$$$ such that $$$1 \le k \le 6$$$, let $$$S$$$ be the set of all indices of the array that are divisible by $$$k$$$.

- When $$$k = 1$$$, $$$S = \{1, 2, 3,4,5,6\}$$$ meaning $$$a_1+a_2+a_3+a_4+a_5+a_6=242$$$ must be divisible by $$$1$$$.
- When $$$k = 2$$$, $$$S = \{2,4,6\}$$$ meaning $$$a_2+a_4+a_6=92$$$ must be divisible by $$$2$$$.
- When $$$k = 3$$$, $$$S = \{3,6\}$$$ meaning $$$a_3+a_6=69$$$ must divisible by $$$3$$$.
- When $$$k = 4$$$, $$$S = \{4\}$$$ meaning $$$a_4=32$$$ must divisible by $$$4$$$.
- When $$$k = 5$$$, $$$S = \{5\}$$$ meaning $$$a_5=125$$$ must divisible by $$$5$$$.
- When $$$k = 6$$$, $$$S = \{6\}$$$ meaning $$$a_6=54$$$ must divisible by $$$6$$$.

Buggy Code:
#include <iostream>
using namespace std ;

int main()
{
    int test ;
    cin >> test ;
    int tarr[test] ;
    for(int l = 0 ; l < test ; l++)
    {
        cin >> tarr[l];
    }
    for(int z = 0 ; z < test ; z++)
    {
        int n = tarr[z];
        int arr[n] = {0} ;
        for(int i = 0 ; i < n ; i++)
        {
            arr[i] = 1 ;
        }
        for(int i = 0 ; i < n ; i++)
        {
            int k = i+1 ;
            int alpha = 1 ;
            for(int j = k ; j <= n ;j = j*alpha)
            {
                arr[j-1] = arr[j-1]*k ;
                alpha++ ;
            }

        }
        for(int i = 0 ; i < n ; i++)
        {
            if(i < n-1)
            {
                cout << arr[i] << " " ;
            }
            else if(i == n-1)
            {
                cout << arr[i] ;
            }

        }
        if(z < test-1)
        {
            cout << '\n' ;
        }

    }

    return 0 ;

}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
1
2
4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Array Divisibility
time_limit_ms: 1000
memory_limit_mb: 256
Description: An array of integers $$$a_1,a_2,\cdots,a_n$$$ is beautiful subject to an integer $$$k$$$ if it satisfies the following:

- The sum of $$$a_{j}$$$ over all $$$j$$$ such that $$$j$$$ is a multiple of $$$k$$$ and $$$1 \le j \le n $$$, itself, is a multiple of $$$k$$$.
- More formally, if $$$\sum_{k | j} a_{j}$$$ is divisible by $$$k$$$ for all $$$1 \le j \le n$$$ then the array $$$a$$$ is beautiful subject to $$$k$$$. Here, the notation $$${k|j}$$$ means $$$k$$$ divides $$$j$$$, that is, $$$j$$$ is a multiple of $$$k$$$.

It can be shown that an answer always exists.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 100$$$). The description of the test cases follows.

The first and only line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$) — the size of the array.

Output: For each test case, print the required array as described in the problem statement.

Examples:
input:
3
3
6
7
output:
4 22 18
10 6 15 32 125 54
23 18 27 36 5 66 7

Note: In the second test case, when $$$n = 6$$$, for all integers $$$k$$$ such that $$$1 \le k \le 6$$$, let $$$S$$$ be the set of all indices of the array that are divisible by $$$k$$$.

- When $$$k = 1$$$, $$$S = \{1, 2, 3,4,5,6\}$$$ meaning $$$a_1+a_2+a_3+a_4+a_5+a_6=242$$$ must be divisible by $$$1$$$.
- When $$$k = 2$$$, $$$S = \{2,4,6\}$$$ meaning $$$a_2+a_4+a_6=92$$$ must be divisible by $$$2$$$.
- When $$$k = 3$$$, $$$S = \{3,6\}$$$ meaning $$$a_3+a_6=69$$$ must divisible by $$$3$$$.
- When $$$k = 4$$$, $$$S = \{4\}$$$ meaning $$$a_4=32$$$ must divisible by $$$4$$$.
- When $$$k = 5$$$, $$$S = \{5\}$$$ meaning $$$a_5=125$$$ must divisible by $$$5$$$.
- When $$$k = 6$$$, $$$S = \{6\}$$$ meaning $$$a_6=54$$$ must divisible by $$$6$$$.

Buggy Code:
#include <iostream>
#include <vector>
#include <map>
#include <algorithm>

using namespace std;

int main() {
  int t;
  cin >> t;

  for (int _ = 0; _ < t; _++) {
    int n;
    cin >> n;

    vector<long long> a(n, 1);

    for (int k = 1; k <= n; k++) {
      for (int i = 0; i < n; i++) {
	if ((i + 1) % k == 0) {
	  a[i] *= k;
	}
      }
    }

    for (int i : a) {
      cout << i << " ";
    }
    cout << "\n";
  }

  return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, let's analyze the code and the problem requirements.

### Problem Analysis:
- The array is considered beautiful if, for every integer \( k \) (where \( 1 \le k \le n \)), the sum of elements at indices that are multiples of \( k \) is divisible by \( k \).
- The buggy code multiplies the elements at indices that are multiples of \( k \) by \( k \) for each \( k \) from 1 to \( n \).

### Bug Analysis:
- The bug in the code is that it multiplies the elements at each step, which can lead to very large values and may not satisfy the condition for all \( k \).

### Test Input Strategy:
- We need a test case where the array size \( n \) is small but still exposes the issue.
- A good strategy is to use a small \( n \) and ensure that the sum of elements at indices that are multiples of \( k \) is not divisible by \( k \) when the buggy code is applied.

### Chosen Test Input:
- Let's use \( n = 4 \) and see if the buggy code fails to produce a valid array.
- For \( n = 4 \), the indices are 1, 2, 3, and 4.
- The buggy code will multiply elements at these indices by 1, 2, 3, and 4 respectively.

### Expected Behavior:
- The buggy code will produce an array where each element is multiplied by its index.
- For example, if the initial array is \([1, 1, 1, 1]\), after applying the buggy code, it will become \([1, 2, 3, 4]\).

### Validation:
- For \( k = 1 \): \( a_1 + a_2 + a_3 + a_4 = 1 + 2 + 3 + 4 = 10 \) (not divisible by 1)
- For \( k = 2 \): \( a_2 + a_4 = 2 + 4 = 6 \) (divisible by 2)
- For \( k = 3 \): \( a_3 = 3 \) (not divisible by 3)
- For \( k = 4 \): \( a_4 = 4 \) (divisible by 4)

### Test Input:
```plaintext
1
4
```

This test input will expose the bug in the buggy code because the sum of elements at indices that are multiples of 3 (which is 3) is not divisible by 3.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Array Divisibility
time_limit_ms: 1000
memory_limit_mb: 256
Description: An array of integers $$$a_1,a_2,\cdots,a_n$$$ is beautiful subject to an integer $$$k$$$ if it satisfies the following:

- The sum of $$$a_{j}$$$ over all $$$j$$$ such that $$$j$$$ is a multiple of $$$k$$$ and $$$1 \le j \le n $$$, itself, is a multiple of $$$k$$$.
- More formally, if $$$\sum_{k | j} a_{j}$$$ is divisible by $$$k$$$ for all $$$1 \le j \le n$$$ then the array $$$a$$$ is beautiful subject to $$$k$$$. Here, the notation $$${k|j}$$$ means $$$k$$$ divides $$$j$$$, that is, $$$j$$$ is a multiple of $$$k$$$.

It can be shown that an answer always exists.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 100$$$). The description of the test cases follows.

The first and only line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$) — the size of the array.

Output: For each test case, print the required array as described in the problem statement.

Examples:
input:
3
3
6
7
output:
4 22 18
10 6 15 32 125 54
23 18 27 36 5 66 7

Note: In the second test case, when $$$n = 6$$$, for all integers $$$k$$$ such that $$$1 \le k \le 6$$$, let $$$S$$$ be the set of all indices of the array that are divisible by $$$k$$$.

- When $$$k = 1$$$, $$$S = \{1, 2, 3,4,5,6\}$$$ meaning $$$a_1+a_2+a_3+a_4+a_5+a_6=242$$$ must be divisible by $$$1$$$.
- When $$$k = 2$$$, $$$S = \{2,4,6\}$$$ meaning $$$a_2+a_4+a_6=92$$$ must be divisible by $$$2$$$.
- When $$$k = 3$$$, $$$S = \{3,6\}$$$ meaning $$$a_3+a_6=69$$$ must divisible by $$$3$$$.
- When $$$k = 4$$$, $$$S = \{4\}$$$ meaning $$$a_4=32$$$ must divisible by $$$4$$$.
- When $$$k = 5$$$, $$$S = \{5\}$$$ meaning $$$a_5=125$$$ must divisible by $$$5$$$.
- When $$$k = 6$$$, $$$S = \{6\}$$$ meaning $$$a_6=54$$$ must divisible by $$$6$$$.

Buggy Code:
def generate_beautiful_array(n):
    # 创建一个长度为 n 的数组，初始值都设为 1
    a = [1] * n

    # 遍历每个 k，从 1 到 n
    for k in range(1, n + 1):
        # 计算需要加到哪些位置上的值，使得这些位置的和对 k 取余为 0
        # 我们选择将这些位置上的值设为 k 的倍数
        for j in range(k - 1, n, k):
            a[j] *= k

    return a

# 读取测试用例的数量
t = int(input())

# 对于每个测试用例
for _ in range(t):
    # 读取数组的大小 n
    n = int(input())

    # 生成美丽的数组
    beautiful_array = generate_beautiful_array(n)

    # 输出结果
    print(' '.join(map(str, beautiful_array)))

# 示例代码的讲解：
# 1. 我们首先定义了一个函数 generate_beautiful_array 来生成美丽的数组。
# 2. 在这个函数中，我们首先创建了一个长度为 n 的数组，初始值都设为 1。
# 3. 然后，我们遍历每个 k，从 1 到 n，对于每个 k，我们找到所有索引 j，使得 j 是 k 的倍数（除了 0）。
# 4. 对于这些索引 j，我们将数组中的对应位置乘以 k，这样就能保证这些位置的和对 k 取余为 0。
# 5. 最后，我们返回生成的数组。
# 6. 在主函数中，我们读取测试用例的数量，然后对于每个测试用例，我们读取 n，生成美丽的数组，并输出。

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the buggy code, we need to consider the following:

1. **Edge Cases**: Test with the smallest and largest possible values for `n`.
2. **Special Cases**: Test with values of `n` that have specific properties, such as being prime or having many divisors.
3. **Intermediate Cases**: Test with values that are neither too small nor too large to ensure the algorithm handles general cases correctly.

Given the constraints:
- \(1 \le t \le 100\)
- \(1 \le n \le 100\)

We will create a test input with a few test cases that cover the above scenarios.

### Test Input
```plaintext
5
1
2
4
6
100
```

### Explanation
1. **n = 1**: This is the smallest possible value for `n`. It should be straightforward, but it's a good edge case to check.
2. **n = 2**: A small value that has a few divisors. This will help check how the algorithm handles simple cases.
3. **n = 4**: A value with a few divisors, including 2 and 4. This will help check how the algorithm handles multiple divisors.
4. **n = 6**: A value with more divisors, including 2, 3, and 6. This will help check how the algorithm handles more complex cases.
5. **n = 100**: The largest possible value for `n`. This will help check how the algorithm handles the upper limit and large arrays.

This test input should help expose any issues in the buggy code, especially related to how it handles different divisors and large arrays.