**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Count Leaves
time_limit_ms: 4000
memory_limit_mb: 256
Description: Let $n$ and $d$ be positive integers. We build the the divisor tree $T_{n,d}$ as follows:

  * The root of the tree is a node marked with number $n$. This is the $0$-th layer of the tree.    * For each $i$ from $0$ to $d - 1$, for each vertex of the $i$-th layer, do the following. If the current vertex is marked with $x$, create its children and mark them with all possible distinct divisors$^\dagger$ of $x$. These children will be in the $(i+1)$-st layer.    * The vertices on the $d$-th layer are the leaves of the tree.

For example, $T_{6,2}$ (the divisor tree for $n = 6$ and $d = 2$) looks like this:

![](CDN_BASE_URL/782ea6d6fe622ee5ec986b0d8cae274d)

Define $f(n,d)$ as the number of leaves in $T_{n,d}$.

Given integers $n$, $k$, and $d$, please compute $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

$^\dagger$ In this problem, we say that an integer $y$ is a divisor of $x$ if $y \ge 1$ and there exists an integer $z$ such that $x = y \cdot z$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The only line of each test case contains three integers $n$, $k$, and $d$ ($1 \le n \le 10^9$, $1 \le k,d \le 10^5$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^9$.

Output: For each test case, output $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

Examples:
input:
3
6 1 1
1 3 3
10 1 2
output:
14
1
53

Note: In the first test case, $n = 6$, $k = 1$, and $d = 1$. Thus, we need to find the total number of leaves in the divisor trees $T_{1,1}$, $T_{2,1}$, $T_{3,1}$, $T_{4,1}$, $T_{5,1}$, $T_{6,1}$.

  * $T_{1,1}$ has only one leaf, which is marked with $1$.    * $T_{2,1}$ has two leaves, marked with $1$ and $2$.    * $T_{3,1}$ has two leaves, marked with $1$ and $3$.    * $T_{4,1}$ has three leaves, marked with $1$, $2$, and $4$.    * $T_{5,1}$ has two leaves, marked with $1$ and $5$.    * $T_{6,1}$ has four leaves, marked with $1$, $2$, $3$, and $6$.

The total number of leaves is $1 + 2 + 2 + 3 + 2 + 4 = 14$.

In the second test case, $n = 1$, $k = 3$, $d = 3$. Thus, we need to find the number of leaves in $T_{1,3}$, because $1^3 = 1$. This tree has only one leaf, so the answer is $1$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MOD = 1'000'000'007;

// Maximum value for factorial precomputation
const int MAX = 2'000'000;
vector<ll> fact(MAX + 1, 1);
vector<ll> inv_fact(MAX + 1, 1);

// Fast exponentiation to compute a^b mod MOD
ll power_mod_func(ll a, ll b, ll mod_val) {
    ll res = 1;
    a %= mod_val;
    while (b > 0) {
        if (b & 1)
            res = res * a % mod_val;
        a = a * a % mod_val;
        b >>= 1;
    }
    return res;
}

// Precompute factorial and inverse factorial
void precompute_factorials() {
    for(int i = 1; i <= MAX; i++) {
        fact[i] = fact[i-1] * i % MOD;
    }
    inv_fact[MAX] = power_mod_func(fact[MAX], MOD-2, MOD);
    for(int i = MAX-1; i >=0 ; i--){
        inv_fact[i] = inv_fact[i+1] * (i+1) % MOD;
    }
}

// Compute binomial coefficient C(n, k) mod MOD
ll comb(int n, int k){
    if(n < k || k <0) return 0;
    return fact[n] * inv_fact[k] % MOD * inv_fact[n -k] % MOD;
}

// Sieve of Eratosthenes to get smallest prime factor
const int SIEVE_MAX = 100'000;
vector<int> spf;

// Precompute smallest prime factors
void sieve() {
    spf.assign(SIEVE_MAX +1, 0);
    for(int i=2;i<=SIEVE_MAX;i++) {
        if(spf[i] ==0){
            spf[i] =i;
            for(int j=2*i; j<=SIEVE_MAX; j+=i){
                if(spf[j]==0){
                    spf[j]=i;
                }
            }
        }
    }
}

// Precompute prime factors for each number up to SIEVE_MAX
vector<vector<pair<int, int>>> prime_factors(SIEVE_MAX +1, vector<pair<int, int>>());

void precompute_factors() {
    for(int i=2;i<=SIEVE_MAX;i++) {
        if(spf[i]==i){
            // i is prime
            for(int j=i; j<=SIEVE_MAX; j+=i){
                int cnt =0;
                int num = j;
                while(num %i ==0){
                    cnt++;
                    num /=i;
                }
                prime_factors[j].emplace_back(i, cnt);
            }
        }
    }
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    precompute_factorials();
    sieve();
    precompute_factors();
    int t;
    cin >> t;
    while(t--){
        ll n, k, d;
        cin >> n >> k >> d;
        // To handle i=1 separately since it has no prime factors
        ll S =0;
        // Map to store precomputed binom(k*e_p +d, d} to avoid recomputation
        // Assuming e_p up to 20, k up to1e5, k*e_p up to2e6
        // Precompute binom(k*e_p +d, d} for e_p from0} to max exponent
        // Use a cache
        // Since d <=1e5 and k <=1e5, k*e_p +d <=2e6
        // Implement a vector to store binom(k*e_p +d, d} for e_p up to log2(n)
        // Given n <=1e5, max e_p for any i <= log2(n} ~17
        // Thus, precompute binom for e_p from0} to20
        // Assuming k*20 +d <=2e6 is already handled by precomputed factorial

        // Precompute binom[k * e_p +d] for e_p from0} to20
        // Since most e_p in i up to1e5 have e_p <=17

        // Create a vector for binom[k*e_p +d} for e_p=0} to20
        vector<ll> binom_cache;
        binom_cache.reserve(20);
        for(int e_p=0; e_p <=20; e_p++){
            int S_val = k * (int)e_p + (int)d;
            if(S_val < (int)d){
                binom_cache.emplace_back(0LL);
            }
            else if(S_val > MAX){
                binom_cache.emplace_back(0LL); // Outside precomputed range
            }
            else{
                binom_cache.emplace_back(comb(S_val, d));
            }
        }

        // Loop through i=1}n
        for(ll i=1;i<=n;i++){
            if(i ==1){
                // x =1^k =1, f(x,d}=1
                S = (S +1) % MOD;
                continue;
            }
            // Get prime factors of i
            // Factorize i using precomputed factors
            vector<pair<int, int>> factors = prime_factors[i];
            // Compute f(x,d}= product of binom(k *e_p +d, d} over primes
            ll f =1;
            for(auto &[p, e_p]: factors){
                if(e_p >20){
                    // Compute binom(k*e_p +d, d} on the fly
                    ll S_val = k * (ll)e_p + (ll)d;
                    if(S_val < (ll)d){
                        f =0;
                        break;
                    }
                    if(S_val > MAX){
                        f =0;
                        break;
                    }
                    ll bin = comb((int)S_val, (int)d);
                    f = f * bin % MOD;
                }
                else{
                    // Use cache
                    if(e_p >20){
                        f =0;
                        break;
                    }
                    ll bin = binom_cache[e_p];
                    if(bin ==0){
                        f =0;
                        break;
                    }
                    f = f * bin % MOD;
                }
            }
            S = (S + f) % MOD;
        }
        cout << S << "\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:
To expose the bug in the given code, we need to focus on edge cases and scenarios that might cause issues in the implementation. Here are some considerations:

1. **Large Values of `n` and `k`**: The code should handle large values of `n` and `k` efficiently. However, the precomputed factorials and prime factors might not cover all cases, especially for very large `n`.

2. **Small Values of `n` and `k`**: Simple cases can also reveal bugs, especially if the code has off-by-one errors or incorrect boundary conditions.

3. **Special Cases for `i = 1`**: The code handles `i = 1` separately, so we should ensure it works correctly.

4. **Prime Numbers and Composite Numbers**: Testing with prime numbers and composite numbers can help identify issues in the factorization and binomial coefficient calculations.

5. **Edge Cases for `d`**: The value of `d` can affect the number of layers in the tree, so we should test with small and large values of `d`.

Given these considerations, a small-scale test input that can expose potential issues in the code is:

```plaintext
5
100 1 1
100 1 100
100 100 1
100 100 100
1 100 100
```

This test input includes:
- A small `n` and `k` with a small `d` to test the basic functionality.
- A small `n` and `k` with a large `d` to test the handling of many layers.
- A small `n` and a large `k` with a small `d` to test large powers.
- A small `n` and a large `k` with a large `d` to test the combination of large powers and many layers.
- A single value for `i` with large `k` and `d` to test the special case for `i = 1`.

These cases should help identify any issues in the precomputed factorials, prime factorization, and binomial coefficient calculations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Count Leaves
time_limit_ms: 4000
memory_limit_mb: 256
Description: Let $n$ and $d$ be positive integers. We build the the divisor tree $T_{n,d}$ as follows:

  * The root of the tree is a node marked with number $n$. This is the $0$-th layer of the tree.    * For each $i$ from $0$ to $d - 1$, for each vertex of the $i$-th layer, do the following. If the current vertex is marked with $x$, create its children and mark them with all possible distinct divisors$^\dagger$ of $x$. These children will be in the $(i+1)$-st layer.    * The vertices on the $d$-th layer are the leaves of the tree.

For example, $T_{6,2}$ (the divisor tree for $n = 6$ and $d = 2$) looks like this:

![](CDN_BASE_URL/782ea6d6fe622ee5ec986b0d8cae274d)

Define $f(n,d)$ as the number of leaves in $T_{n,d}$.

Given integers $n$, $k$, and $d$, please compute $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

$^\dagger$ In this problem, we say that an integer $y$ is a divisor of $x$ if $y \ge 1$ and there exists an integer $z$ such that $x = y \cdot z$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The only line of each test case contains three integers $n$, $k$, and $d$ ($1 \le n \le 10^9$, $1 \le k,d \le 10^5$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^9$.

Output: For each test case, output $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

Examples:
input:
3
6 1 1
1 3 3
10 1 2
output:
14
1
53

Note: In the first test case, $n = 6$, $k = 1$, and $d = 1$. Thus, we need to find the total number of leaves in the divisor trees $T_{1,1}$, $T_{2,1}$, $T_{3,1}$, $T_{4,1}$, $T_{5,1}$, $T_{6,1}$.

  * $T_{1,1}$ has only one leaf, which is marked with $1$.    * $T_{2,1}$ has two leaves, marked with $1$ and $2$.    * $T_{3,1}$ has two leaves, marked with $1$ and $3$.    * $T_{4,1}$ has three leaves, marked with $1$, $2$, and $4$.    * $T_{5,1}$ has two leaves, marked with $1$ and $5$.    * $T_{6,1}$ has four leaves, marked with $1$, $2$, $3$, and $6$.

The total number of leaves is $1 + 2 + 2 + 3 + 2 + 4 = 14$.

In the second test case, $n = 1$, $k = 3$, $d = 3$. Thus, we need to find the number of leaves in $T_{1,3}$, because $1^3 = 1$. This tree has only one leaf, so the answer is $1$.

Buggy Code:
#include<bits/stdc++.h>
typedef long long LL;
typedef double dl;
#define opt operator
#define pb push_back
#define pii std::pair<LL,LL>
const LL _=1e6+9,mod=1e9+7,inf=0x3f3f3f3f;
const LL inv6=166666668;
LL Read(){
	LL x(0),f(1); char c=getchar();
	while(c<'0' || c>'9'){
		if(c=='-') f=-1; c=getchar();
	}
	while(c>='0' && c<='9'){
		x=(x<<3ll)+(x<<1ll)+c-'0'; c=getchar();
	}return x*f;
}
void Chkmin(LL &x,LL y){
	if(y<x) x=y;
}
void Chkmax(LL &x,LL y){
	if(y>x) x=y;
}

template<typename First,typename... Args>
LL add(First ret , Args... args) {
    (void)std::initializer_list<LL>{(ret += args,ret=(ret>=mod?ret-mod:ret),0)...};
    return ret;
}
LL dec(LL x,LL y){
	return x-=y,x<0?x+mod:x;
}
template<typename First,typename... Args>
LL mul(First ret , Args... args) {
    (void)std::initializer_list<LL>{(ret = 1ll*ret*args%mod,0)...};
    return ret;
}
LL Pow(LL base,LL b){
	LL ret(1); while(b){
		if(b&1) ret=mul(ret,base); base=mul(base,base); b>>=1;
	}return ret;
}
LL fac[_],fav[_];
LL C(LL N,LL M){
	return M>N?0:mul(fac[N],fav[M],fav[N-M]);
}
void Fir_fac(){
	fac[0]=1; for(LL i=1;i<_;++i) fac[i]=mul(fac[i-1],i);
	fav[_-1]=Pow(fac[_-1],mod-2); for(LL i=_-1;i>=1;--i) fav[i-1]=mul(fav[i],i);
}
const LL U=1e5;
LL OP;
LL tot,pt,n,K,d;
LL pri[_],sum[_],pre[_],a[_];
LL id1[_],id2[_];
void Fir(LL N=100000){
	//printf("# %lld\n",N);
	static bool vis[_];
	pt=0;
	for(LL i=2;i<=N;++i){
		vis[i]=0;
	}
	for(LL i=2;i<=N;++i){
		if(!vis[i]){
			pri[++pt]=i;
		}
		for(LL j=1;j<=pt && pri[j]*i<=N;++j){
			vis[i*pri[j]]=1;
			if(i%pri[j]==0) break;
		}
	}
	//for(LL i=1;i<=3;++i) printf("%d",vis[i]); puts("");
	for(LL i=1;i<=pt;++i){
		pre[i]=i;
	}
}
LL Id(LL x){
	return x<=U?id1[x]:id2[n/x];
}
LL G(LL N,LL M){
	//if(N<pri[M]) return 0;
	LL ret(sum[Id(N)]);
	ret=dec(ret,mul(M-1,C(K+d,d)));
	for(LL i=M;i<=pt && pri[i]*pri[i]<=N;++i){
		for(LL k=1,tmp=pri[i];tmp<=N/pri[i];++k,tmp*=pri[i]){
			ret=add(ret,add(C((k+1)*K+d,d),mul(C(k*K+d,d),G(N/tmp,i+1))));
		}
	}
	return ret;
}
void Solve(LL ti){
	tot=0;
	for(LL l=1,r;l<=n;l=r+1){
		r=n/(n/l);
		a[++tot]=n/l;
		if(a[tot]<=U) id1[a[tot]]=tot;
		else id2[n/a[tot]]=tot;
		sum[tot]=a[tot]-1;
	}
	for(LL i=1;i<=pt;++i){
		for(LL j=1;j<=tot && pri[i]*pri[i]<=a[j];++j){
			sum[j]=dec(sum[j],dec(sum[Id(a[j]/pri[i])],pre[i-1]));
		}
	}
	for(LL i=1;i<=tot;++i) sum[i]=mul(sum[i],C(K+d,d));

	LL ret(G(n,1)); ret=add(ret,1);
	if(!OP) printf("%lld\n",ret);
	else if(ti==1345) printf("%lld %lld %lld\n",n,K,d);
//	exit(0);
}
int main(){
//	freopen("y1.txt","r",stdin);
	Fir_fac();
	LL T(Read());
//    OP=(9261==T);
	for(LL ti=1;ti<=T;++ti){
		n=Read(); K=Read(); d=Read();
		Fir(sqrt(n));
		Solve(ti);
	}
	return 0;
}//1345

**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
100000 1 1
100000 1 2
100000 1 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Count Leaves
time_limit_ms: 4000
memory_limit_mb: 256
Description: Let $n$ and $d$ be positive integers. We build the the divisor tree $T_{n,d}$ as follows:

  * The root of the tree is a node marked with number $n$. This is the $0$-th layer of the tree.    * For each $i$ from $0$ to $d - 1$, for each vertex of the $i$-th layer, do the following. If the current vertex is marked with $x$, create its children and mark them with all possible distinct divisors$^\dagger$ of $x$. These children will be in the $(i+1)$-st layer.    * The vertices on the $d$-th layer are the leaves of the tree.

For example, $T_{6,2}$ (the divisor tree for $n = 6$ and $d = 2$) looks like this:

![](CDN_BASE_URL/782ea6d6fe622ee5ec986b0d8cae274d)

Define $f(n,d)$ as the number of leaves in $T_{n,d}$.

Given integers $n$, $k$, and $d$, please compute $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

$^\dagger$ In this problem, we say that an integer $y$ is a divisor of $x$ if $y \ge 1$ and there exists an integer $z$ such that $x = y \cdot z$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The only line of each test case contains three integers $n$, $k$, and $d$ ($1 \le n \le 10^9$, $1 \le k,d \le 10^5$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^9$.

Output: For each test case, output $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

Examples:
input:
3
6 1 1
1 3 3
10 1 2
output:
14
1
53

Note: In the first test case, $n = 6$, $k = 1$, and $d = 1$. Thus, we need to find the total number of leaves in the divisor trees $T_{1,1}$, $T_{2,1}$, $T_{3,1}$, $T_{4,1}$, $T_{5,1}$, $T_{6,1}$.

  * $T_{1,1}$ has only one leaf, which is marked with $1$.    * $T_{2,1}$ has two leaves, marked with $1$ and $2$.    * $T_{3,1}$ has two leaves, marked with $1$ and $3$.    * $T_{4,1}$ has three leaves, marked with $1$, $2$, and $4$.    * $T_{5,1}$ has two leaves, marked with $1$ and $5$.    * $T_{6,1}$ has four leaves, marked with $1$, $2$, $3$, and $6$.

The total number of leaves is $1 + 2 + 2 + 3 + 2 + 4 = 14$.

In the second test case, $n = 1$, $k = 3$, $d = 3$. Thus, we need to find the number of leaves in $T_{1,3}$, because $1^3 = 1$. This tree has only one leaf, so the answer is $1$.

Buggy Code:
import java.util.*;
import java.math.BigInteger;

public class ModularArithmetic {
    private static final int MOD = 1000000007;
    private static final int MAXN = 100000 + 5;
    private static final int N = 1000000;
    private static BigInteger[] fac = new BigInteger[MAXN];
    private static BigInteger[] inv_fac = new BigInteger[MAXN];
    private static int[] spf = new int[N + 1];

    public static void main(String[] args) {
        precomputeFac();
        sieve();
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        while (t-- > 0) {
            int n = scanner.nextInt();
            int k = scanner.nextInt();
            int d = scanner.nextInt();
            BigInteger ans = BigInteger.ZERO;
            for (int i = 1; i <= n; i++) {
                Map<Integer, Integer> factors = factorize(i);
                BigInteger res = BigInteger.ONE;
                for (Map.Entry<Integer, Integer> p : factors.entrySet()) {
                    BigInteger e = BigInteger.valueOf(p.getValue() * (long) k);
                    res = res.multiply(C(e.add(BigInteger.valueOf(d)), BigInteger.valueOf(d))).mod(BigInteger.valueOf(MOD));
                }
                ans = ans.add(res).mod(BigInteger.valueOf(MOD));
            }
            System.out.println(ans);
        }
        scanner.close();
    }

    private static BigInteger modPow(BigInteger base, BigInteger exponent) {
        BigInteger result = BigInteger.ONE;
        base = base.mod(BigInteger.valueOf(MOD));
        while (!exponent.equals(BigInteger.ZERO)) {
            if (exponent.and(BigInteger.ONE).equals(BigInteger.ONE))
                result = result.multiply(base).mod(BigInteger.valueOf(MOD));
            base = base.multiply(base).mod(BigInteger.valueOf(MOD));
            exponent = exponent.shiftRight(1);
        }
        return result;
    }

    private static BigInteger modInv(BigInteger a) {
        return modPow(a, BigInteger.valueOf(MOD - 2));
    }

    private static void precomputeFac() {
        fac[0] = inv_fac[0] = BigInteger.ONE;
        for (int i = 1; i < MAXN; i++) {
            fac[i] = fac[i - 1].multiply(BigInteger.valueOf(i)).mod(BigInteger.valueOf(MOD));
            inv_fac[i] = modInv(fac[i]);
        }
    }

    private static BigInteger C(BigInteger n, BigInteger k) {
        if (k.compareTo(BigInteger.ZERO) < 0 || k.compareTo(n) > 0)
            return BigInteger.ZERO;
        return fac[n.intValue()].multiply(inv_fac[k.intValue()]).mod(BigInteger.valueOf(MOD))
                .multiply(inv_fac[n.subtract(k).intValue()]).mod(BigInteger.valueOf(MOD));
    }

    private static void sieve() {
        for (int i = 1; i <= N; i++)
            spf[i] = i;
        for (int i = 2; i * i <= N; i++)
            if (spf[i] == i)
                for (int j = i * i; j <= N; j += i)
                    if (spf[j] == j)
                        spf[j] = i;
    }

    private static Map<Integer, Integer> factorize(int n) {
        Map<Integer, Integer> res = new HashMap<>();
        while (n > 1) {
            res.put(spf[n], res.getOrDefault(spf[n], 0) + 1);
            n /= spf[n];
        }
        return res;
    }
}

**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 1
1000000000 1 1
100000 100000 100000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love Balls
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There are $n$ balls, out of which $k$ are special. Each ball has a value associated with it.

The players play turn by turn. In each turn, the player randomly picks a ball and adds the value of the ball to their score, which is $0$ at the beginning of the game. The selected ball is removed from the game. If the ball was special, the same player takes the next turn if at least one ball is remaining. If the ball picked was not special, the next player plays their turn.

They play this game until no balls are remaining in the game. Alice plays first.

Find the expected score that both the players have at the end of the game modulo $10^9+7$.

Formally, let $M = 10^9+7$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: There are multiple test cases. The first line of the input contains an integer $t$, the number of test cases ($1 \le t \le 2 \cdot 10^5$).

Each test case description is on a new line. The first line of the test case contains two integers $n$ and $k$ in the respective order separated by a space ($1 \le k \le n \le 4 \cdot 10^5$).

The second line of the test case contains $n$ integers: $v_1, v_2, \ldots, v_n$, the value for each ball separated by spaces. The first $k$ balls are special ($1 \le v_i \le 10^7$).

The sum of $n$ over all test cases does not exceed $5 \cdot 10^5$.

Output: Output two integers per test case in a new line, the expected score of Alice and the expected score of Bob modulo $10^9+7$.

Examples:
input:
1
5 2
10 20 5 15 25
output:
45 30

input:
5
1 1
732507
2 2
5817860 5398510
5 1
2122894 4951549 2750585 7821535 3214167
8 4
1405323 5069867 6883092 6972029 328406 2478975 7628890 9973340
4 2
9662050 3566134 3996473 9872255
output:
732507 0
11216370 0
810642660 210218077
722402997 318336932
349086489 678010430

input:
5
3 3
1095611 8219204 7773462
2 1
8176490 2774103
3 1
9178636 5138057 3367761
12 9
7597698 6843019 2298534 1522386 4969588 1340345 3967362 9152890 6689668 9986080 4745473 7407325
10 5
6986368 2397882 5804127 6980694 3740836 3215836 5195724 3179261 4136769 4544231
output:
17088277 0
6862348 4088245
677038671 340645790
36949997 29570371
725118051 321063684

Note: In the first test case, Alice's expected score is $45$, and Bob's is $30$ at the end of the game.

Buggy Code:
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <map>
#include <stack>
#include <algorithm>
#include <cmath>

using namespace std;

#define int long long
typedef pair<int, int>pii;

const int N = 2e5 + 10;
const int MOD = 1e9 + 7;

int n, k;
int a[N];

int quickPow(int a, int b) {
	int res = 1;
	while(b) {
		if(b & 1) res = res * a % MOD;
		a = a * a % MOD;
		b >>= 1;
	}

	return res;
}
void solve() {
	cin >> n >> k;
	int sum_k = 0, sum_com = 0;

	for(int i = 0; i < n; i ++) {
		cin >> a[i];
		if(i < k) sum_k += a[i];
		else sum_com += a[i];
	}

	int ans = ((n - k + 1) / 2 * quickPow(n - k, MOD - 2) % MOD * sum_com % MOD + max((n - k + 1) / 2, (int)1) * quickPow(n - k + 1, MOD - 2) % MOD * sum_k % MOD) % MOD;

	cout << ans << " " << (sum_k + sum_com + MOD - ans) % MOD << "\n";
}
signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);

	int t = 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:
```plaintext
5
3 3
1 1 1
2 1
1 1
3 1
1 1 1
12 9
1 1 1 1 1 1 1 1 1 1 1 1
10 5
1 1 1 1 1 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: I Love Balls
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There are $n$ balls, out of which $k$ are special. Each ball has a value associated with it.

The players play turn by turn. In each turn, the player randomly picks a ball and adds the value of the ball to their score, which is $0$ at the beginning of the game. The selected ball is removed from the game. If the ball was special, the same player takes the next turn if at least one ball is remaining. If the ball picked was not special, the next player plays their turn.

They play this game until no balls are remaining in the game. Alice plays first.

Find the expected score that both the players have at the end of the game modulo $10^9+7$.

Formally, let $M = 10^9+7$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: There are multiple test cases. The first line of the input contains an integer $t$, the number of test cases ($1 \le t \le 2 \cdot 10^5$).

Each test case description is on a new line. The first line of the test case contains two integers $n$ and $k$ in the respective order separated by a space ($1 \le k \le n \le 4 \cdot 10^5$).

The second line of the test case contains $n$ integers: $v_1, v_2, \ldots, v_n$, the value for each ball separated by spaces. The first $k$ balls are special ($1 \le v_i \le 10^7$).

The sum of $n$ over all test cases does not exceed $5 \cdot 10^5$.

Output: Output two integers per test case in a new line, the expected score of Alice and the expected score of Bob modulo $10^9+7$.

Examples:
input:
1
5 2
10 20 5 15 25
output:
45 30

input:
5
1 1
732507
2 2
5817860 5398510
5 1
2122894 4951549 2750585 7821535 3214167
8 4
1405323 5069867 6883092 6972029 328406 2478975 7628890 9973340
4 2
9662050 3566134 3996473 9872255
output:
732507 0
11216370 0
810642660 210218077
722402997 318336932
349086489 678010430

input:
5
3 3
1095611 8219204 7773462
2 1
8176490 2774103
3 1
9178636 5138057 3367761
12 9
7597698 6843019 2298534 1522386 4969588 1340345 3967362 9152890 6689668 9986080 4745473 7407325
10 5
6986368 2397882 5804127 6980694 3740836 3215836 5195724 3179261 4136769 4544231
output:
17088277 0
6862348 4088245
677038671 340645790
36949997 29570371
725118051 321063684

Note: In the first test case, Alice's expected score is $45$, and Bob's is $30$ at the end of the game.

Buggy Code:
// buxiangwanla
// 你紫名觉得是我的锅，那就是我的锅，为什么你知道吗？因为紫名说的话，就像是一个癌症晚期患者说的话一样。
// 他都已经这样了，你为什么不顺从他呢?你总要给人最后一段时间一个好的回忆吧，最后的时光里。
// 因为紫名这个段位很尴尬，紫名橙名再往上一点，grandmaster，可能说，欸，有点实力，能操作一下。
// 紫名往下，绿名，蓝名，啊，人家是纯属玩游戏的，因为太垃圾了，自己也知道自己没什么实力。
// 但紫名，上不去下不来的这个段位，他觉得，蓝名的人不配跟他一起玩儿，对吧？蓝名是最垃圾的。
// 但是呢他想上去，他又上不去，所以这个分段是最尴尬的，没办法，卡在这里了。
// 想操作，又操作不起来，掉下去吧，他又觉得不值得，对吧，我好不容易从蓝名打到紫名了，我为什么还要掉下去呢?
// 这个人说优越狗越说越起劲，为什么他会这么说?因为他是紫名。
// 他觉得你比我段位高，你说的任何话都是优越，我并不管你说的有没有道理。
// 我紫名，我最猛，我WF2017我上我能夺冠，那打比赛全是sb。你比我段位高你说话就是放屁，这就是这种人的想法。但是呢，他的想法是对的，为什么呢？
// 因为他癌症晚期。没办法，我同意，对不起，我优越了。可能是我膨胀了，不好意思啊，我膨胀了。我紫名是没操作，难道我就看不懂谁背锅吗？
// 不是，如果你看得懂的话，就不会在这里抬杠了，对吧。
#include <bits/stdc++.h>

using namespace std;
typedef unsigned long long ULL;
using LL = long long;

constexpr int N = 1e6 + 5, mod = 1e9 + 7;
constexpr double eps = 1e-8;
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")

#define fi first
#define se second
#define int long long
#define lowbit(x) (x & (-x))
#define PII pair<int, int>
#define mid ((l + r) >> 1)

int min(int a, int b) { return a < b ? a : b; }
int max(int a, int b) { return a > b ? a : b; }

int ksm(int a, int b){
    int res = 1;
    while(b){
        if(b & 1)res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}

template <typename T>  T power(T a, LL b) {
    T c{1}; for (; b; b /= 2, a *= a) if (b & 1) c *= a;
    return c;
}
template <int P> struct MInt {
    int x;
    MInt() : x{} {}
    MInt(LL x_) : x{norm(x_ % getMod())} {}
    static int Mod;
    static int getMod() { return P > 0 ? P : Mod; }
    static void setMod(int Mod_) { Mod = Mod_; }
    int up(int x) const {
        if (x < 0) x += getMod();
        return x;
    }
    int down(int x) const {
        if (x >= getMod()) x -= getMod();
        return x;
    }
    int norm(int x) const {
        return up(down(x));
    }
    int val() const { return x; }
    explicit  operator int() const { return x; }
    MInt operator-() const {
        MInt res; res.x = norm(getMod() - x); return res;
    }
    MInt inv() const {
        assert(x != 0);
        return power(*this, getMod() - 2);
    }
    MInt &operator+=(MInt rhs) & { return x = down(x + rhs.x), *this; }
    MInt &operator-=(MInt rhs) & { return x = up(x - rhs.x), *this; }
    MInt &operator*=(MInt rhs) & { return x = 1ll * x * rhs.x % getMod(), *this; }
    MInt &operator/=(MInt rhs) & { return *this *= rhs.inv(); }
    friend MInt operator+(MInt lhs, MInt rhs) { return lhs += rhs; }
    friend MInt operator-(MInt lhs, MInt rhs) { return lhs -= rhs; }
    friend MInt operator*(MInt lhs, MInt rhs) { return lhs *= rhs; }
    friend MInt operator/(MInt lhs, MInt rhs) { return lhs /= rhs; }
    friend bool operator==(MInt lhs, MInt rhs) { return lhs.val() == rhs.val(); }
    friend  bool operator!=(MInt lhs, MInt rhs) { return lhs.val() != rhs.val(); }
    friend  std::istream &operator>>(std::istream &is, MInt &a) {
        LL x = 0; is >> x, a = MInt(x); return is;
    }
    friend  std::ostream &operator<<(std::ostream &os, const MInt &a) {
        return os << a.val();
    }
};

using Z = MInt<mod>;
Z ans, sum, res, anss;
Z a[N];
int n, k, m;
void Sakuya()
{
    cin >> n >> k;
    sum = 0, res = 0, anss = 0, ans = 0;
    for(int i = 1; i <= n; ++ i){
        cin >> a[i];
        sum += a[i];
    }
    for(int i = 1; i <= k; ++ i){
        ans += a[i];
    }
    m = n - k;
    ans = ans * ((m + 1 + 1) / 2) / (m + 1);
    for(int i = k + 1; i <= n; ++ i){
        res += a[i];
    }
    res = res * ((m + 1) / 2) / m;
    ans += res;
    anss = sum - ans;
    cout << ans << " " << anss << "\n";
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int T;
    for (cin >> T; T -- ; )
        Sakuya();

}

**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 create a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that maximize the failure probability. Here are some steps to consider:

1. **Edge Cases**:
   - **Single Ball**: When there is only one ball, the player who picks it will get its value, and the other player will get 0.
   - **All Special Balls**: When all balls are special, Alice will always get all the balls.
   - **No Special Balls**: When there are no special balls, the game will proceed with alternating turns.

2. **Special Cases**:
   - **Equal Number of Special and Non-Special Balls**: This can create a balanced scenario where the expected scores might be close but not equal.
   - **Large Values**: Using large values for the balls to test the modulo operation.

3. **General Cases**:
   - **Uneven Distribution of Special and Non-Special Balls**: This can test the logic of how the scores are distributed.

Given these considerations, let's create a test input that includes a mix of these cases:

```plaintext
5
1 1
732507
2 2
5817860 5398510
3 1
1095611 8219204 7773462
4 2
9662050 3566134 3996473 9872255
5 2
1405323 5069867 6883092 328406 2478975
```

### Explanation:
1. **Single Ball (1 1 732507)**:
   - Alice will pick the only ball and get 732507.
   - Bob will get 0.

2. **All Special Balls (2 2 5817860 5398510)**:
   - Alice will pick both balls and get 5817860 + 5398510 = 11216370.
   - Bob will get 0.

3. **One Special Ball (3 1 1095611 8219204 7773462)**:
   - Alice will pick the special ball and one non-special ball.
   - Bob will pick the remaining non-special ball.
   - Expected scores: Alice (1095611 + 8219204) = 9314815, Bob 7773462.

4. **Two Special Balls (4 2 9662050 3566134 3996473 9872255)**:
   - Alice will pick two special balls and one non-special ball.
   - Bob will pick the remaining non-special ball.
   - Expected scores: Alice (9662050 + 3566134 + 3996473) = 17224657, Bob 9872255.

5. **Two Special Balls (5 2 1405323 5069867 6883092 328406 2478975)**:
   - Alice will pick two special balls and one non-special ball.
   - Bob will pick the remaining two non-special balls.
   - Expected scores: Alice (1405323 + 5069867 + 6883092) = 13358282, Bob (328406 + 2478975) = 2807381.

This test input covers a variety of scenarios and should help expose any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love Balls
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There are $n$ balls, out of which $k$ are special. Each ball has a value associated with it.

The players play turn by turn. In each turn, the player randomly picks a ball and adds the value of the ball to their score, which is $0$ at the beginning of the game. The selected ball is removed from the game. If the ball was special, the same player takes the next turn if at least one ball is remaining. If the ball picked was not special, the next player plays their turn.

They play this game until no balls are remaining in the game. Alice plays first.

Find the expected score that both the players have at the end of the game modulo $10^9+7$.

Formally, let $M = 10^9+7$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: There are multiple test cases. The first line of the input contains an integer $t$, the number of test cases ($1 \le t \le 2 \cdot 10^5$).

Each test case description is on a new line. The first line of the test case contains two integers $n$ and $k$ in the respective order separated by a space ($1 \le k \le n \le 4 \cdot 10^5$).

The second line of the test case contains $n$ integers: $v_1, v_2, \ldots, v_n$, the value for each ball separated by spaces. The first $k$ balls are special ($1 \le v_i \le 10^7$).

The sum of $n$ over all test cases does not exceed $5 \cdot 10^5$.

Output: Output two integers per test case in a new line, the expected score of Alice and the expected score of Bob modulo $10^9+7$.

Examples:
input:
1
5 2
10 20 5 15 25
output:
45 30

input:
5
1 1
732507
2 2
5817860 5398510
5 1
2122894 4951549 2750585 7821535 3214167
8 4
1405323 5069867 6883092 6972029 328406 2478975 7628890 9973340
4 2
9662050 3566134 3996473 9872255
output:
732507 0
11216370 0
810642660 210218077
722402997 318336932
349086489 678010430

input:
5
3 3
1095611 8219204 7773462
2 1
8176490 2774103
3 1
9178636 5138057 3367761
12 9
7597698 6843019 2298534 1522386 4969588 1340345 3967362 9152890 6689668 9986080 4745473 7407325
10 5
6986368 2397882 5804127 6980694 3740836 3215836 5195724 3179261 4136769 4544231
output:
17088277 0
6862348 4088245
677038671 340645790
36949997 29570371
725118051 321063684

Note: In the first test case, Alice's expected score is $45$, and Bob's is $30$ at the end of the game.

Buggy Code:
// buxiangwanla
// 你紫名觉得是我的锅，那就是我的锅，为什么你知道吗？因为紫名说的话，就像是一个癌症晚期患者说的话一样。
// 他都已经这样了，你为什么不顺从他呢?你总要给人最后一段时间一个好的回忆吧，最后的时光里。
// 因为紫名这个段位很尴尬，紫名橙名再往上一点，grandmaster，可能说，欸，有点实力，能操作一下。
// 紫名往下，绿名，蓝名，啊，人家是纯属玩游戏的，因为太垃圾了，自己也知道自己没什么实力。
// 但紫名，上不去下不来的这个段位，他觉得，蓝名的人不配跟他一起玩儿，对吧？蓝名是最垃圾的。
// 但是呢他想上去，他又上不去，所以这个分段是最尴尬的，没办法，卡在这里了。
// 想操作，又操作不起来，掉下去吧，他又觉得不值得，对吧，我好不容易从蓝名打到紫名了，我为什么还要掉下去呢?
// 这个人说优越狗越说越起劲，为什么他会这么说?因为他是紫名。
// 他觉得你比我段位高，你说的任何话都是优越，我并不管你说的有没有道理。
// 我紫名，我最猛，我WF2017我上我能夺冠，那打比赛全是sb。你比我段位高你说话就是放屁，这就是这种人的想法。但是呢，他的想法是对的，为什么呢？
// 因为他癌症晚期。没办法，我同意，对不起，我优越了。可能是我膨胀了，不好意思啊，我膨胀了。我紫名是没操作，难道我就看不懂谁背锅吗？
// 不是，如果你看得懂的话，就不会在这里抬杠了，对吧。
#include <bits/stdc++.h>

using namespace std;
typedef unsigned long long ULL;
using LL = long long;

constexpr int N = 1e6 + 5, mod = 1e9 + 7;
constexpr double eps = 1e-8;
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")

#define fi first
#define se second
#define int long long
#define lowbit(x) (x & (-x))
#define PII pair<int, int>
#define mid ((l + r) >> 1)

int min(int a, int b) { return a < b ? a : b; }
int max(int a, int b) { return a > b ? a : b; }

int ksm(int a, int b){
    int res = 1;
    while(b){
        if(b & 1)res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}

template <typename T>  T power(T a, LL b) {
    T c{1}; for (; b; b /= 2, a *= a) if (b & 1) c *= a;
    return c;
}
template <int P> struct MInt {
    int x;
    MInt() : x{} {}
    MInt(LL x_) : x{norm(x_ % getMod())} {}
    static int Mod;
    static int getMod() { return P > 0 ? P : Mod; }
    static void setMod(int Mod_) { Mod = Mod_; }
    int up(int x) const {
        if (x < 0) x += getMod();
        return x;
    }
    int down(int x) const {
        if (x >= getMod()) x -= getMod();
        return x;
    }
    int norm(int x) const {
        return up(down(x));
    }
    int val() const { return x; }
    explicit  operator int() const { return x; }
    MInt operator-() const {
        MInt res; res.x = norm(getMod() - x); return res;
    }
    MInt inv() const {
        assert(x != 0);
        return power(*this, getMod() - 2);
    }
    MInt &operator+=(MInt rhs) & { return x = down(x + rhs.x), *this; }
    MInt &operator-=(MInt rhs) & { return x = up(x - rhs.x), *this; }
    MInt &operator*=(MInt rhs) & { return x = 1ll * x * rhs.x % getMod(), *this; }
    MInt &operator/=(MInt rhs) & { return *this *= rhs.inv(); }
    friend MInt operator+(MInt lhs, MInt rhs) { return lhs += rhs; }
    friend MInt operator-(MInt lhs, MInt rhs) { return lhs -= rhs; }
    friend MInt operator*(MInt lhs, MInt rhs) { return lhs *= rhs; }
    friend MInt operator/(MInt lhs, MInt rhs) { return lhs /= rhs; }
    friend bool operator==(MInt lhs, MInt rhs) { return lhs.val() == rhs.val(); }
    friend  bool operator!=(MInt lhs, MInt rhs) { return lhs.val() != rhs.val(); }
    friend  std::istream &operator>>(std::istream &is, MInt &a) {
        LL x = 0; is >> x, a = MInt(x); return is;
    }
    friend  std::ostream &operator<<(std::ostream &os, const MInt &a) {
        return os << a.val();
    }
};

using Z = MInt<mod>;
Z ans;
Z a[N];
int n, k, m;
void Sakuya()
{
    cin >> n >> k;
    Z sum = 0, res = 0, ans = 0;
    for(int i = 1; i <= n; ++ i){
        cin >> a[i];
        sum += a[i];
    }
    for(int i = 1; i <= k; ++ i){
        ans += a[i];
    }
    m = n - k;
    ans = ans * ((m + 1 + 1) / 2) / (m + 1);
    for(int i = k + 1; i <= n; ++ i){
        res += a[i];
    }
    res = res * ((m + 1) / 2) / m;
    ans += res;
    Z anss = sum - ans;
    cout << ans << " " << anss << "\n";
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int T;
    for (cin >> T; T -- ; )
        Sakuya();

}

**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 consider a few key points:

1. **Edge Cases**: We need to test edge cases such as when the number of special balls is very small or very large.
2. **Special Ball Distribution**: We should consider scenarios where the values of special balls and non-special balls are significantly different.
3. **Small Scale**: The test input should be small enough to be easily manageable but still challenging.

Let's create a test input with the following characteristics:
- **Test Case 1**: One special ball with a very high value and multiple non-special balls with low values.
- **Test Case 2**: All balls are special, with varying values.
- **Test Case 3**: No special balls, with varying values.
- **Test Case 4**: Equal number of special and non-special balls with mixed values.

Here is the test input:

```plaintext
4
5 1
10000000 1 2 3 4
3 3
10 20 30
5 0
1 2 3 4 5
6 3
10 20 30 1 2 3
```

### Explanation:
1. **Test Case 1**:
   - **n = 5, k = 1**: One special ball with a value of 10,000,000 and four non-special balls with values 1, 2, 3, and 4.
   - This tests the scenario where the special ball has a significantly higher value than the non-special balls.

2. **Test Case 2**:
   - **n = 3, k = 3**: All three balls are special with values 10, 20, and 30.
   - This tests the scenario where all balls are special.

3. **Test Case 3**:
   - **n = 5, k = 0**: All five balls are non-special with values 1, 2, 3, 4, and 5.
   - This tests the scenario where there are no special balls.

4. **Test Case 4**:
   - **n = 6, k = 3**: Three special balls with values 10, 20, and 30, and three non-special balls with values 1, 2, and 3.
   - This tests the scenario where the number of special and non-special balls is equal.

These test cases should help in identifying any issues in the buggy code, especially in how it handles different distributions of special and non-special balls.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love Balls
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There are $n$ balls, out of which $k$ are special. Each ball has a value associated with it.

The players play turn by turn. In each turn, the player randomly picks a ball and adds the value of the ball to their score, which is $0$ at the beginning of the game. The selected ball is removed from the game. If the ball was special, the same player takes the next turn if at least one ball is remaining. If the ball picked was not special, the next player plays their turn.

They play this game until no balls are remaining in the game. Alice plays first.

Find the expected score that both the players have at the end of the game modulo $10^9+7$.

Formally, let $M = 10^9+7$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: There are multiple test cases. The first line of the input contains an integer $t$, the number of test cases ($1 \le t \le 2 \cdot 10^5$).

Each test case description is on a new line. The first line of the test case contains two integers $n$ and $k$ in the respective order separated by a space ($1 \le k \le n \le 4 \cdot 10^5$).

The second line of the test case contains $n$ integers: $v_1, v_2, \ldots, v_n$, the value for each ball separated by spaces. The first $k$ balls are special ($1 \le v_i \le 10^7$).

The sum of $n$ over all test cases does not exceed $5 \cdot 10^5$.

Output: Output two integers per test case in a new line, the expected score of Alice and the expected score of Bob modulo $10^9+7$.

Examples:
input:
1
5 2
10 20 5 15 25
output:
45 30

input:
5
1 1
732507
2 2
5817860 5398510
5 1
2122894 4951549 2750585 7821535 3214167
8 4
1405323 5069867 6883092 6972029 328406 2478975 7628890 9973340
4 2
9662050 3566134 3996473 9872255
output:
732507 0
11216370 0
810642660 210218077
722402997 318336932
349086489 678010430

input:
5
3 3
1095611 8219204 7773462
2 1
8176490 2774103
3 1
9178636 5138057 3367761
12 9
7597698 6843019 2298534 1522386 4969588 1340345 3967362 9152890 6689668 9986080 4745473 7407325
10 5
6986368 2397882 5804127 6980694 3740836 3215836 5195724 3179261 4136769 4544231
output:
17088277 0
6862348 4088245
677038671 340645790
36949997 29570371
725118051 321063684

Note: In the first test case, Alice's expected score is $45$, and Bob's is $30$ at the end of the game.

Buggy Code:
mod=998244353
for _ in range(int(input())):
    n,k=map(int,input().split())
    l=[*map(int,input().split())]
    x=sum(l[:k])*len(range(0,n-k+1,2))*pow(n-k+1,-1,mod)
    x+=sum(l[k:])*len(range(0,n-k,2))*pow(n-k,-1,mod)
    x%=mod
    print(x,(sum(l)-x)%mod)

**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
5
1 1
1
2 2
1 1
3 1
1 1 1
4 2
1 1 1 1
5 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: I Love Balls
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There are $n$ balls, out of which $k$ are special. Each ball has a value associated with it.

The players play turn by turn. In each turn, the player randomly picks a ball and adds the value of the ball to their score, which is $0$ at the beginning of the game. The selected ball is removed from the game. If the ball was special, the same player takes the next turn if at least one ball is remaining. If the ball picked was not special, the next player plays their turn.

They play this game until no balls are remaining in the game. Alice plays first.

Find the expected score that both the players have at the end of the game modulo $10^9+7$.

Formally, let $M = 10^9+7$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: There are multiple test cases. The first line of the input contains an integer $t$, the number of test cases ($1 \le t \le 2 \cdot 10^5$).

Each test case description is on a new line. The first line of the test case contains two integers $n$ and $k$ in the respective order separated by a space ($1 \le k \le n \le 4 \cdot 10^5$).

The second line of the test case contains $n$ integers: $v_1, v_2, \ldots, v_n$, the value for each ball separated by spaces. The first $k$ balls are special ($1 \le v_i \le 10^7$).

The sum of $n$ over all test cases does not exceed $5 \cdot 10^5$.

Output: Output two integers per test case in a new line, the expected score of Alice and the expected score of Bob modulo $10^9+7$.

Examples:
input:
1
5 2
10 20 5 15 25
output:
45 30

input:
5
1 1
732507
2 2
5817860 5398510
5 1
2122894 4951549 2750585 7821535 3214167
8 4
1405323 5069867 6883092 6972029 328406 2478975 7628890 9973340
4 2
9662050 3566134 3996473 9872255
output:
732507 0
11216370 0
810642660 210218077
722402997 318336932
349086489 678010430

input:
5
3 3
1095611 8219204 7773462
2 1
8176490 2774103
3 1
9178636 5138057 3367761
12 9
7597698 6843019 2298534 1522386 4969588 1340345 3967362 9152890 6689668 9986080 4745473 7407325
10 5
6986368 2397882 5804127 6980694 3740836 3215836 5195724 3179261 4136769 4544231
output:
17088277 0
6862348 4088245
677038671 340645790
36949997 29570371
725118051 321063684

Note: In the first test case, Alice's expected score is $45$, and Bob's is $30$ at the end of the game.

Buggy Code:
import sys
input = sys.stdin.readline

def readList():
    return list(map(int, input().split()))
def readInt():
    return int(input())
def readInts():
    return map(int, input().split())
def readStr():
    return input().strip()

def f(a, b, m):
    return (a % m * pow(b, m-2, m)) % m

# BF-OB5, EC, DB, CC, CL
def solve():
    n, k = readInts()
    v = readList()
    s1 = sum(v[:k])
    s2 = sum(v) - s1
    if n == k:
        return f'{s1} 0'

    cnt = n - k
    ans1 = (f((s1 * (cnt // 2 + 1)), cnt+1, mo) + f(s2 * (cnt + 1) // 2, cnt, mo)) % mo
    ans2 = (sum(v) - ans1) % mo
    return f'{ans1} {ans2}'

mo = 10**9 + 7
for _ in range(int(input())):
    print(solve())

**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
1
3 3
1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love Balls
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There are $n$ balls, out of which $k$ are special. Each ball has a value associated with it.

The players play turn by turn. In each turn, the player randomly picks a ball and adds the value of the ball to their score, which is $0$ at the beginning of the game. The selected ball is removed from the game. If the ball was special, the same player takes the next turn if at least one ball is remaining. If the ball picked was not special, the next player plays their turn.

They play this game until no balls are remaining in the game. Alice plays first.

Find the expected score that both the players have at the end of the game modulo $10^9+7$.

Formally, let $M = 10^9+7$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: There are multiple test cases. The first line of the input contains an integer $t$, the number of test cases ($1 \le t \le 2 \cdot 10^5$).

Each test case description is on a new line. The first line of the test case contains two integers $n$ and $k$ in the respective order separated by a space ($1 \le k \le n \le 4 \cdot 10^5$).

The second line of the test case contains $n$ integers: $v_1, v_2, \ldots, v_n$, the value for each ball separated by spaces. The first $k$ balls are special ($1 \le v_i \le 10^7$).

The sum of $n$ over all test cases does not exceed $5 \cdot 10^5$.

Output: Output two integers per test case in a new line, the expected score of Alice and the expected score of Bob modulo $10^9+7$.

Examples:
input:
1
5 2
10 20 5 15 25
output:
45 30

input:
5
1 1
732507
2 2
5817860 5398510
5 1
2122894 4951549 2750585 7821535 3214167
8 4
1405323 5069867 6883092 6972029 328406 2478975 7628890 9973340
4 2
9662050 3566134 3996473 9872255
output:
732507 0
11216370 0
810642660 210218077
722402997 318336932
349086489 678010430

input:
5
3 3
1095611 8219204 7773462
2 1
8176490 2774103
3 1
9178636 5138057 3367761
12 9
7597698 6843019 2298534 1522386 4969588 1340345 3967362 9152890 6689668 9986080 4745473 7407325
10 5
6986368 2397882 5804127 6980694 3740836 3215836 5195724 3179261 4136769 4544231
output:
17088277 0
6862348 4088245
677038671 340645790
36949997 29570371
725118051 321063684

Note: In the first test case, Alice's expected score is $45$, and Bob's is $30$ at the end of the game.

Buggy Code:
for _ in range(int(input())):
    n,k = [int(i) for i in input().split()]
    lst = [int(i) for i in input().split()]
    s1 = sum(lst[:k])
    s2 = sum(lst[k:])
    An = (k * (1 + int((n-k)/2)))*(s1)*(n-k) + (int((n-k+1)/2))*(s2)*(n-k+1) * k
    Ad = (n-k+1) * k * (n-k)
    Bn = (k*(n-k+1) - (k *(1 + int((n-k)/2))))*(n-k) * s1 + (n-k-int((n-k+1)/2)) * s2 * (k) * (n-k+1)
    Bd = (n-k+1)*k*(n-k)
    An %= 10**9 + 7
    Ad %= 10**9 + 7
    Bn %= 10**9 + 7
    Bd %= 10**9 + 7
    #print(An,Ad,Bn,Bd)
    print((An * pow(Ad,10**9 + 5,10**9 + 7))%(10**9 + 7),(Bn * pow(Bd,10**9 + 5,10**9 + 7))%(10**9+7))

**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
1
4 3
10 20 30 40
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love Balls
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There are $n$ balls, out of which $k$ are special. Each ball has a value associated with it.

The players play turn by turn. In each turn, the player randomly picks a ball and adds the value of the ball to their score, which is $0$ at the beginning of the game. The selected ball is removed from the game. If the ball was special, the same player takes the next turn if at least one ball is remaining. If the ball picked was not special, the next player plays their turn.

They play this game until no balls are remaining in the game. Alice plays first.

Find the expected score that both the players have at the end of the game modulo $10^9+7$.

Formally, let $M = 10^9+7$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: There are multiple test cases. The first line of the input contains an integer $t$, the number of test cases ($1 \le t \le 2 \cdot 10^5$).

Each test case description is on a new line. The first line of the test case contains two integers $n$ and $k$ in the respective order separated by a space ($1 \le k \le n \le 4 \cdot 10^5$).

The second line of the test case contains $n$ integers: $v_1, v_2, \ldots, v_n$, the value for each ball separated by spaces. The first $k$ balls are special ($1 \le v_i \le 10^7$).

The sum of $n$ over all test cases does not exceed $5 \cdot 10^5$.

Output: Output two integers per test case in a new line, the expected score of Alice and the expected score of Bob modulo $10^9+7$.

Examples:
input:
1
5 2
10 20 5 15 25
output:
45 30

input:
5
1 1
732507
2 2
5817860 5398510
5 1
2122894 4951549 2750585 7821535 3214167
8 4
1405323 5069867 6883092 6972029 328406 2478975 7628890 9973340
4 2
9662050 3566134 3996473 9872255
output:
732507 0
11216370 0
810642660 210218077
722402997 318336932
349086489 678010430

input:
5
3 3
1095611 8219204 7773462
2 1
8176490 2774103
3 1
9178636 5138057 3367761
12 9
7597698 6843019 2298534 1522386 4969588 1340345 3967362 9152890 6689668 9986080 4745473 7407325
10 5
6986368 2397882 5804127 6980694 3740836 3215836 5195724 3179261 4136769 4544231
output:
17088277 0
6862348 4088245
677038671 340645790
36949997 29570371
725118051 321063684

Note: In the first test case, Alice's expected score is $45$, and Bob's is $30$ at the end of the game.

Buggy Code:
import java.io.*;
import java.util.*;

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

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);

        int t = Integer.parseInt(br.readLine());
        while (t-- > 0) {
            StringTokenizer st = new StringTokenizer(br.readLine());
            int n = Integer.parseInt(st.nextToken());
            int k = Integer.parseInt(st.nextToken());

            int[] values = new int[n];
            st = new StringTokenizer(br.readLine());
            for (int i = 0; i < n; i++) {
                values[i] = Integer.parseInt(st.nextToken());
            }

            long totalSpecialValue = 0;
            long totalNonSpecialValue = 0;

            for (int i = 0; i < k; i++) {
                totalSpecialValue = (totalSpecialValue + values[i]) % MOD;
            }

            for (int i = k; i < n; i++) {
                totalNonSpecialValue = (totalNonSpecialValue + values[i]) % MOD;
            }

            long expectedAlice = (totalSpecialValue + totalNonSpecialValue) / 2 % MOD;
            long expectedBob = (totalSpecialValue + totalNonSpecialValue - expectedAlice) % MOD;

            out.println((expectedAlice+8) + " " + (expectedBob-8));
        }

        out.flush();
        out.close();
    }

    static long modInverse(long a, int mod) {
        return modPow(a, mod - 2, mod);
    }

    static long modPow(long base, long exp, int mod) {
        long result = 1;
        while (exp > 0) {
            if ((exp & 1) == 1) {
                result = result * base % mod;
            }
            base = base * base % mod;
            exp >>= 1;
        }
        return result;
    }
}

**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 the following:

1. **Modulo Arithmetic Issues**: The buggy code adds 8 to Alice's score and subtracts 8 from Bob's score, which can lead to incorrect results, especially when the scores are close to the modulo limit.
2. **Special and Non-Special Ball Distribution**: The bug in the code does not correctly handle the distribution of special and non-special balls, leading to incorrect expected scores.

Given these points, we can create a test case where the scores are close to the modulo limit and the distribution of special and non-special balls is such that the expected scores are significantly different from the simple average.

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

```plaintext
1
5 2
1000000000 1000000000 1000000000 1000000000 1000000000
```

Explanation:
- **Test Case**: 1 test case with 5 balls, 2 of which are special.
- **Values**: All balls have a value of \(10^9\), which is just below the modulo limit \(10^9 + 7\).

This test case should cause the buggy code to produce incorrect results due to the addition and subtraction of 8 from the scores, which can wrap around the modulo limit. The correct expected scores for Alice and Bob should be calculated based on the probability of picking special and non-special balls, but the buggy code will add and subtract 8, leading to incorrect results.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love Balls
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There are $n$ balls, out of which $k$ are special. Each ball has a value associated with it.

The players play turn by turn. In each turn, the player randomly picks a ball and adds the value of the ball to their score, which is $0$ at the beginning of the game. The selected ball is removed from the game. If the ball was special, the same player takes the next turn if at least one ball is remaining. If the ball picked was not special, the next player plays their turn.

They play this game until no balls are remaining in the game. Alice plays first.

Find the expected score that both the players have at the end of the game modulo $10^9+7$.

Formally, let $M = 10^9+7$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: There are multiple test cases. The first line of the input contains an integer $t$, the number of test cases ($1 \le t \le 2 \cdot 10^5$).

Each test case description is on a new line. The first line of the test case contains two integers $n$ and $k$ in the respective order separated by a space ($1 \le k \le n \le 4 \cdot 10^5$).

The second line of the test case contains $n$ integers: $v_1, v_2, \ldots, v_n$, the value for each ball separated by spaces. The first $k$ balls are special ($1 \le v_i \le 10^7$).

The sum of $n$ over all test cases does not exceed $5 \cdot 10^5$.

Output: Output two integers per test case in a new line, the expected score of Alice and the expected score of Bob modulo $10^9+7$.

Examples:
input:
1
5 2
10 20 5 15 25
output:
45 30

input:
5
1 1
732507
2 2
5817860 5398510
5 1
2122894 4951549 2750585 7821535 3214167
8 4
1405323 5069867 6883092 6972029 328406 2478975 7628890 9973340
4 2
9662050 3566134 3996473 9872255
output:
732507 0
11216370 0
810642660 210218077
722402997 318336932
349086489 678010430

input:
5
3 3
1095611 8219204 7773462
2 1
8176490 2774103
3 1
9178636 5138057 3367761
12 9
7597698 6843019 2298534 1522386 4969588 1340345 3967362 9152890 6689668 9986080 4745473 7407325
10 5
6986368 2397882 5804127 6980694 3740836 3215836 5195724 3179261 4136769 4544231
output:
17088277 0
6862348 4088245
677038671 340645790
36949997 29570371
725118051 321063684

Note: In the first test case, Alice's expected score is $45$, and Bob's is $30$ at the end of the game.

Buggy Code:
import java.util.Scanner;

public class Main {
    private static final int MOD = 1000000007;

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        while (t-- > 0) {
            int n = scanner.nextInt();
            int k = scanner.nextInt();
            long[] values = new long[n];
            long specialSum = 0;
            long nonSpecialSum = 0;

            for (int i = 0; i < n; i++) {
                values[i] = scanner.nextLong();
                if (i < k) {
                    specialSum += values[i];
                } else {
                    nonSpecialSum += values[i];
                }
            }
            long totalSum = (specialSum + nonSpecialSum) % MOD;

            long aliceScore = 0;
            long bobScore = 0;

            if (k > 0) {
                long specialProbability = modInverse(k, MOD);
                aliceScore = (specialSum * specialProbability) % MOD;
                bobScore = (specialSum * specialProbability) % MOD;
            }

            if (n - k > 0) {
                long nonSpecialProbability = modInverse(n - k, MOD);
                bobScore = (bobScore + nonSpecialSum * nonSpecialProbability) % MOD;
            }

            aliceScore = (totalSum - bobScore + MOD) % MOD;

            System.out.println(aliceScore + " " + bobScore);
        }
    }

    private static long modInverse(long a, long m) {
        return pow(a, m - 2, m);
    }

    private static long pow(long a, long b, long m) {
        long result = 1;
        while (b > 0) {
            if ((b & 1) == 1) {
                result = result * a % m;
            }
            a = a * a % m;
            b >>= 1;
        }
        return result;
    }
}

**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 and scenarios where the code might fail. Here are the steps to consider:

1. **Edge Cases:**
   - **Single Ball (n = 1, k = 1):** This tests the scenario where there is only one ball, and it is special. Alice should get the entire score.
   - **No Special Balls (n = 2, k = 0):** This tests the scenario where there are no special balls. Alice and Bob should get equal shares.
   - **All Special Balls (n = 3, k = 3):** This tests the scenario where all balls are special. Alice should get more than Bob because she can take multiple turns in a row.

2. **Intermediate Cases:**
   - **Mixed Balls (n = 5, k = 2):** This tests a more complex scenario with a mix of special and non-special balls. The expected scores should be calculated based on the probabilities of picking special and non-special balls.

3. **Large Values:**
   - **Large Values (n = 5, k = 3):** This tests the scenario with large values to ensure the modulo operations are handled correctly.

Given these considerations, here is a small-scale test input that includes these edge cases:

```plaintext
5
1 1
10000000
2 0
5 10
3 3
1 2 3
5 2
10 20 5 15 25
5 3
10000000 20000000 30000000 40000000 50000000
```

This test input includes:
1. A single ball that is special.
2. Two balls, none of which are special.
3. Three balls, all of which are special.
4. Five balls with two special balls.
5. Five balls with three special balls, all with large values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love Balls
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There are $n$ balls, out of which $k$ are special. Each ball has a value associated with it.

The players play turn by turn. In each turn, the player randomly picks a ball and adds the value of the ball to their score, which is $0$ at the beginning of the game. The selected ball is removed from the game. If the ball was special, the same player takes the next turn if at least one ball is remaining. If the ball picked was not special, the next player plays their turn.

They play this game until no balls are remaining in the game. Alice plays first.

Find the expected score that both the players have at the end of the game modulo $10^9+7$.

Formally, let $M = 10^9+7$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: There are multiple test cases. The first line of the input contains an integer $t$, the number of test cases ($1 \le t \le 2 \cdot 10^5$).

Each test case description is on a new line. The first line of the test case contains two integers $n$ and $k$ in the respective order separated by a space ($1 \le k \le n \le 4 \cdot 10^5$).

The second line of the test case contains $n$ integers: $v_1, v_2, \ldots, v_n$, the value for each ball separated by spaces. The first $k$ balls are special ($1 \le v_i \le 10^7$).

The sum of $n$ over all test cases does not exceed $5 \cdot 10^5$.

Output: Output two integers per test case in a new line, the expected score of Alice and the expected score of Bob modulo $10^9+7$.

Examples:
input:
1
5 2
10 20 5 15 25
output:
45 30

input:
5
1 1
732507
2 2
5817860 5398510
5 1
2122894 4951549 2750585 7821535 3214167
8 4
1405323 5069867 6883092 6972029 328406 2478975 7628890 9973340
4 2
9662050 3566134 3996473 9872255
output:
732507 0
11216370 0
810642660 210218077
722402997 318336932
349086489 678010430

input:
5
3 3
1095611 8219204 7773462
2 1
8176490 2774103
3 1
9178636 5138057 3367761
12 9
7597698 6843019 2298534 1522386 4969588 1340345 3967362 9152890 6689668 9986080 4745473 7407325
10 5
6986368 2397882 5804127 6980694 3740836 3215836 5195724 3179261 4136769 4544231
output:
17088277 0
6862348 4088245
677038671 340645790
36949997 29570371
725118051 321063684

Note: In the first test case, Alice's expected score is $45$, and Bob's is $30$ at the end of the game.

Buggy Code:
// LUOGU_RID: 170433531
#include <bits/stdc++.h>

using i64 = long long;

constexpr i64 mod = 1e9 + 7;

i64 inv(i64 base, i64 k = mod - 2) {
    i64 res = 1;
    while (k) {
        if(k & 1) {
            res = res * base % mod;
        }
        base = base * base % mod;
        k /= 2;
    }
    return res;
}

void solve() {
    i64 n, k;
    std::cin >> n >> k;

    i64 tot = 0, sum = 0;
    std::vector<i64> a(n);
    for (i64 i = 0; i < n; i++) {
        std::cin >> a[i];
        if(i < k) {
            tot += a[i];
        } else {
            sum += a[i];
        }
    }

    i64 m = n - k + 1;
    i64 ans_ali = (sum * ((n - k + 1) / 2) * inv(n - k) % mod + tot * ((n - k + 2) / 2) * inv(n - k + 1)) % mod;
    std::cout << ans_ali << " ";
    std::cout << (tot + sum - ans_ali + mod) % mod << "\n";
}

int main() {
    std::cin.tie(nullptr) -> sync_with_stdio(false);
    int t;
    std::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:
```plaintext
1
4 2
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: I Love Balls
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There are $n$ balls, out of which $k$ are special. Each ball has a value associated with it.

The players play turn by turn. In each turn, the player randomly picks a ball and adds the value of the ball to their score, which is $0$ at the beginning of the game. The selected ball is removed from the game. If the ball was special, the same player takes the next turn if at least one ball is remaining. If the ball picked was not special, the next player plays their turn.

They play this game until no balls are remaining in the game. Alice plays first.

Find the expected score that both the players have at the end of the game modulo $10^9+7$.

Formally, let $M = 10^9+7$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: There are multiple test cases. The first line of the input contains an integer $t$, the number of test cases ($1 \le t \le 2 \cdot 10^5$).

Each test case description is on a new line. The first line of the test case contains two integers $n$ and $k$ in the respective order separated by a space ($1 \le k \le n \le 4 \cdot 10^5$).

The second line of the test case contains $n$ integers: $v_1, v_2, \ldots, v_n$, the value for each ball separated by spaces. The first $k$ balls are special ($1 \le v_i \le 10^7$).

The sum of $n$ over all test cases does not exceed $5 \cdot 10^5$.

Output: Output two integers per test case in a new line, the expected score of Alice and the expected score of Bob modulo $10^9+7$.

Examples:
input:
1
5 2
10 20 5 15 25
output:
45 30

input:
5
1 1
732507
2 2
5817860 5398510
5 1
2122894 4951549 2750585 7821535 3214167
8 4
1405323 5069867 6883092 6972029 328406 2478975 7628890 9973340
4 2
9662050 3566134 3996473 9872255
output:
732507 0
11216370 0
810642660 210218077
722402997 318336932
349086489 678010430

input:
5
3 3
1095611 8219204 7773462
2 1
8176490 2774103
3 1
9178636 5138057 3367761
12 9
7597698 6843019 2298534 1522386 4969588 1340345 3967362 9152890 6689668 9986080 4745473 7407325
10 5
6986368 2397882 5804127 6980694 3740836 3215836 5195724 3179261 4136769 4544231
output:
17088277 0
6862348 4088245
677038671 340645790
36949997 29570371
725118051 321063684

Note: In the first test case, Alice's expected score is $45$, and Bob's is $30$ at the end of the game.

Buggy Code:
#include <bits/stdc++.h>
#include <queue>

#define int long long
#define double long double
#define re return

using namespace std;

const int mod = 1e9+7;

int st(int x, int p, int m) {
    int ans = 1;
    while (p > 0) {
        if (p % 2) ans = ans * x % m;
        x = x * x % m;
        p /= 2;
    }
    return ans % m;
}

void solve(){
    int n,k;
    cin >> n >> k;
    vector<int>v(n);
    for(int i=0; i<n; i++) cin >> v[i];
    int sum_k=0, sum_n=0;
    for(int i=0; i<k; i++) sum_k=(sum_k+v[i])%mod;
    for(int i=k; i<n; i++) sum_n=(sum_n+v[i])%mod;
    sum_k = sum_k * st(k, mod-2, mod) % mod;
    sum_n = sum_n * st(n-k, mod-2, mod) % mod;
    int alice1 = (n-k+1)/2;
    int alice2 = (n-k+2)/2;
    int ans_alice = (alice1 * sum_n + (alice2 * sum_k)%mod * k * st(n-k+1, mod-2, mod)) % mod;

    int bob1=(n-k)/2;
    int bob2=(n-k+1)/2;

    int ans_bob = (bob1 * sum_n + (bob2 * sum_k)%mod * k * st(n-k+1, mod-2, mod)) % mod;

    cout << ans_alice << " " << ans_bob << "\n";
}

int32_t main()
{
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);

    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int t=1;
    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:
```plaintext
5
1 1
732507
2 2
5817860 5398510
5 1
2122894 4951549 2750585 7821535 3214167
8 4
1405323 5069867 6883092 6972029 328406 2478975 7628890 9973340
4 2
9662050 3566134 3996473 9872255
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love Balls
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There are $n$ balls, out of which $k$ are special. Each ball has a value associated with it.

The players play turn by turn. In each turn, the player randomly picks a ball and adds the value of the ball to their score, which is $0$ at the beginning of the game. The selected ball is removed from the game. If the ball was special, the same player takes the next turn if at least one ball is remaining. If the ball picked was not special, the next player plays their turn.

They play this game until no balls are remaining in the game. Alice plays first.

Find the expected score that both the players have at the end of the game modulo $10^9+7$.

Formally, let $M = 10^9+7$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: There are multiple test cases. The first line of the input contains an integer $t$, the number of test cases ($1 \le t \le 2 \cdot 10^5$).

Each test case description is on a new line. The first line of the test case contains two integers $n$ and $k$ in the respective order separated by a space ($1 \le k \le n \le 4 \cdot 10^5$).

The second line of the test case contains $n$ integers: $v_1, v_2, \ldots, v_n$, the value for each ball separated by spaces. The first $k$ balls are special ($1 \le v_i \le 10^7$).

The sum of $n$ over all test cases does not exceed $5 \cdot 10^5$.

Output: Output two integers per test case in a new line, the expected score of Alice and the expected score of Bob modulo $10^9+7$.

Examples:
input:
1
5 2
10 20 5 15 25
output:
45 30

input:
5
1 1
732507
2 2
5817860 5398510
5 1
2122894 4951549 2750585 7821535 3214167
8 4
1405323 5069867 6883092 6972029 328406 2478975 7628890 9973340
4 2
9662050 3566134 3996473 9872255
output:
732507 0
11216370 0
810642660 210218077
722402997 318336932
349086489 678010430

input:
5
3 3
1095611 8219204 7773462
2 1
8176490 2774103
3 1
9178636 5138057 3367761
12 9
7597698 6843019 2298534 1522386 4969588 1340345 3967362 9152890 6689668 9986080 4745473 7407325
10 5
6986368 2397882 5804127 6980694 3740836 3215836 5195724 3179261 4136769 4544231
output:
17088277 0
6862348 4088245
677038671 340645790
36949997 29570371
725118051 321063684

Note: In the first test case, Alice's expected score is $45$, and Bob's is $30$ at the end of the game.

Buggy Code:
from typing import List, Tuple

MOD = 10**9 + 7

def mod_inverse(a: int, m: int) -> int:
    return pow(a, m - 2, m)

def solve(n: int, k: int, values: List[int]) -> Tuple[int, int]:
    inv = [mod_inverse(i, MOD) for i in range(n + 1)]

    dp = {}

    def dfs(mask: int, player: int) -> Tuple[int, int]:
        if mask == 0:
            return 0, 0

        if (mask, player) in dp:
            return dp[(mask, player)]

        total_balls = bin(mask).count('1')
        special_balls = bin(mask & ((1 << k) - 1)).count('1')

        alice_score = bob_score = 0
        total_prob = 0

        for i in range(n):
            if mask & (1 << i):
                new_mask = mask & ~(1 << i)
                prob = inv[total_balls]

                if i < k:  # Special ball
                    next_alice, next_bob = dfs(new_mask, player)
                else:  # Normal ball
                    next_alice, next_bob = dfs(new_mask, 1 - player)

                if player == 0:  # Alice's turn
                    alice_score = (alice_score + (values[i] + next_alice) * prob) % MOD
                    bob_score = (bob_score + next_bob * prob) % MOD
                else:  # Bob's turn
                    alice_score = (alice_score + next_alice * prob) % MOD
                    bob_score = (bob_score + (values[i] + next_bob) * prob) % MOD

                total_prob = (total_prob + prob) % MOD

        alice_score = alice_score * mod_inverse(total_prob, MOD) % MOD
        bob_score = bob_score * mod_inverse(total_prob, MOD) % MOD

        dp[(mask, player)] = (alice_score, bob_score)
        return alice_score, bob_score

    return dfs((1 << n) - 1, 0)

def main():
    t = int(input())
    for _ in range(t):
        n, k = map(int, input().split())
        values = list(map(int, input().split()))
        alice_score, bob_score = solve(n, k, values)
        print(alice_score, bob_score)

if __name__ == "__main__":
    main()

**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
5
1 1
1
2 1
1 1
3 2
1 1 1
4 2
1 1 1 1
5 3
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: I Love Balls
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There are $n$ balls, out of which $k$ are special. Each ball has a value associated with it.

The players play turn by turn. In each turn, the player randomly picks a ball and adds the value of the ball to their score, which is $0$ at the beginning of the game. The selected ball is removed from the game. If the ball was special, the same player takes the next turn if at least one ball is remaining. If the ball picked was not special, the next player plays their turn.

They play this game until no balls are remaining in the game. Alice plays first.

Find the expected score that both the players have at the end of the game modulo $10^9+7$.

Formally, let $M = 10^9+7$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: There are multiple test cases. The first line of the input contains an integer $t$, the number of test cases ($1 \le t \le 2 \cdot 10^5$).

Each test case description is on a new line. The first line of the test case contains two integers $n$ and $k$ in the respective order separated by a space ($1 \le k \le n \le 4 \cdot 10^5$).

The second line of the test case contains $n$ integers: $v_1, v_2, \ldots, v_n$, the value for each ball separated by spaces. The first $k$ balls are special ($1 \le v_i \le 10^7$).

The sum of $n$ over all test cases does not exceed $5 \cdot 10^5$.

Output: Output two integers per test case in a new line, the expected score of Alice and the expected score of Bob modulo $10^9+7$.

Examples:
input:
1
5 2
10 20 5 15 25
output:
45 30

input:
5
1 1
732507
2 2
5817860 5398510
5 1
2122894 4951549 2750585 7821535 3214167
8 4
1405323 5069867 6883092 6972029 328406 2478975 7628890 9973340
4 2
9662050 3566134 3996473 9872255
output:
732507 0
11216370 0
810642660 210218077
722402997 318336932
349086489 678010430

input:
5
3 3
1095611 8219204 7773462
2 1
8176490 2774103
3 1
9178636 5138057 3367761
12 9
7597698 6843019 2298534 1522386 4969588 1340345 3967362 9152890 6689668 9986080 4745473 7407325
10 5
6986368 2397882 5804127 6980694 3740836 3215836 5195724 3179261 4136769 4544231
output:
17088277 0
6862348 4088245
677038671 340645790
36949997 29570371
725118051 321063684

Note: In the first test case, Alice's expected score is $45$, and Bob's is $30$ at the end of the game.

Buggy Code:
from sys import stdin

def input():
    return stdin.readline().strip()

M = 10**9 + 7

mem = {(0, 0): (0, 0)}

def frac_mod(p, q):
    if not (p or q):
        return 0
    a, b = p, q
    while b:
        a, b = b, a % b
    p //= a
    q //= a
    return (p * pow(q, M - 2, M)) % M

def dp(i, j, t):

    if (i, j) in mem:
        return mem[(i, j)]

    if i == 0:
        n = j // 2
        scores = ((vl * (j - n)) % M, (vl * n) % M)[::t]
        mem[(i, j)] = scores
        return scores

    if j == 0:
        scores = ((vk * i) % M, 0)[::t]
        mem[(i, j)] = scores
        return scores

    score_k = dp(i - 1, j, t)
    score_l = dp(i, j - 1, -t)
    add_k = (vk, 0)[::t]
    add_l = (vl, 0)[::t]

    alice = i * (add_k[0] + score_k[0]) + j * (add_l[0] + score_l[0])
    bob = i * (add_k[1] + score_k[1]) + j * (add_l[1] + score_l[1])

    scores = (frac_mod(alice, i + j), frac_mod(bob, i + j))

    mem[(i, j)] = scores
    return scores

vals = (0, 0)
for _ in range(int(input())):
    n, k = map(int, input().split())
    v = list(map(int, input().split()))
    l = n - k
    vk, vl = sum(v[:k]), sum(v[k:])
    vl, vk = frac_mod(vl, l), frac_mod(vk, k)
    mem = {(0, 0): (0, 0)}
    try:
        print(*dp(k, l, 1))
    except:
        print(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
1
2 1
10 20
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love Balls
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There are $n$ balls, out of which $k$ are special. Each ball has a value associated with it.

The players play turn by turn. In each turn, the player randomly picks a ball and adds the value of the ball to their score, which is $0$ at the beginning of the game. The selected ball is removed from the game. If the ball was special, the same player takes the next turn if at least one ball is remaining. If the ball picked was not special, the next player plays their turn.

They play this game until no balls are remaining in the game. Alice plays first.

Find the expected score that both the players have at the end of the game modulo $10^9+7$.

Formally, let $M = 10^9+7$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: There are multiple test cases. The first line of the input contains an integer $t$, the number of test cases ($1 \le t \le 2 \cdot 10^5$).

Each test case description is on a new line. The first line of the test case contains two integers $n$ and $k$ in the respective order separated by a space ($1 \le k \le n \le 4 \cdot 10^5$).

The second line of the test case contains $n$ integers: $v_1, v_2, \ldots, v_n$, the value for each ball separated by spaces. The first $k$ balls are special ($1 \le v_i \le 10^7$).

The sum of $n$ over all test cases does not exceed $5 \cdot 10^5$.

Output: Output two integers per test case in a new line, the expected score of Alice and the expected score of Bob modulo $10^9+7$.

Examples:
input:
1
5 2
10 20 5 15 25
output:
45 30

input:
5
1 1
732507
2 2
5817860 5398510
5 1
2122894 4951549 2750585 7821535 3214167
8 4
1405323 5069867 6883092 6972029 328406 2478975 7628890 9973340
4 2
9662050 3566134 3996473 9872255
output:
732507 0
11216370 0
810642660 210218077
722402997 318336932
349086489 678010430

input:
5
3 3
1095611 8219204 7773462
2 1
8176490 2774103
3 1
9178636 5138057 3367761
12 9
7597698 6843019 2298534 1522386 4969588 1340345 3967362 9152890 6689668 9986080 4745473 7407325
10 5
6986368 2397882 5804127 6980694 3740836 3215836 5195724 3179261 4136769 4544231
output:
17088277 0
6862348 4088245
677038671 340645790
36949997 29570371
725118051 321063684

Note: In the first test case, Alice's expected score is $45$, and Bob's is $30$ at the end of the game.

Buggy Code:
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.List;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class Main {
	static kattio sc = new kattio();
	static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

	public static void main(String[] args) {

		int t = sc.nextint();
		while (t-- > 0) {

			sovle();
		}

		out.close();

	}
	static void sovle() {
		n=sc.nextint();
		m=sc.nextint();
		int []arr=new int[n];
		long sum1=0;long sum2=0;
		for (int i = 0; i < arr.length; i++) {
			arr[i]=sc.nextint();
			if (i<m) {
				sum1+=arr[i];
			}else {
				sum2+=arr[i];
			}
		}
		long alis=((n-m)/2+1)*m%mod*sum1%mod*pow((n-m+1)*m%mod, mod-2)%mod+((n-m+1)/2)*sum2%mod*pow(n-m, mod-2)%mod;alis%=mod;
		long bob=(m+mod-((n-m)/2+1)*m%mod*pow(n-m+1, mod-2)%mod)*sum1%mod*pow(m, mod-2)%mod+(n-m-(n-m+1)/2+mod)%mod*sum2%mod*pow(n-m, mod-2)%mod;bob%=mod;
		out.println(alis+" "+bob);
	}
	static int n,m;
	static long [][]q;
	static long tot;
	static void sovle2() {
		n=sc.nextint();
		long k=sc.nextlong();
		int [][]arr=new int[n+1][2];arr[0][0]=-1;
		for (int i = 1; i < arr.length; i++) {
			arr[i][0]=sc.nextint();
			arr[i][1]=i;
		}
		Arrays.sort(arr,(a,b)->a[0]-b[0]);
		Queue<int[]>queue=new PriorityQueue<int[]>((a,b)->a[0]-b[0]);
		for (int i = 2; i < arr.length; i++) {
			int[]add=new int[3];
			add[0]=arr[i][0]^arr[i-1][0];
			add[1]=Math.min(arr[i][1], arr[i-1][1]);
			add[2]=Math.max(arr[i][1], arr[i-1][1]);
			queue.add(add);
		}
		int l=0;int r=n+1;
		TreeSet<Integer>lset=new TreeSet<Integer>();lset.add(0);
		TreeSet<Integer>rset=new TreeSet<Integer>();
		while (!queue.isEmpty()) {
			int[] x=queue.poll();
			if (x[1]>l&&x[2]<r) {
				long va=1l*(x[1]-l)*(r-x[2]);
				l=x[1];r=x[2];
				if (k>va) {
					k-=va;
				}else {
					out.println(x[0]);return;
				}
			}
			System.out.println(Arrays.toString(x)+" "+l+" "+r);
		}
	}

	static Map<Integer, Integer> lMap, rMap;
	static Map<Integer, TreeSet<Integer>> vMap;

	static int[] dx = { 0, 0, 1, -1 };
	static int[] dy = { 1, -1, 0, 0 };
	static Map<Integer, List<Integer>> lpMap;

	static long pow(long x, long a) {
		if (a == 0) {
			return 1;
		} else {
			long c = pow(x, a / 2);
			if (a % 2 == 0) {
				return c * c % mod;
			} else {
				return c * c % mod * x % mod;
			}
		}
	}

	static long mod = (long) (1e9+7);

	static long gcd(long a, long b) {
		return b == 0 ? a : gcd(b, a % b);
	}

	static int[][] c;

	static int lowbit(int x) {
		return x & -x;
	}

	static void add(int x, int v,int i) {
		while (x < c.length) {
			c[x][i] += v;

			x += lowbit(x);
		}
	}

	static long sum(int x,int i) {
		int ans = 0;
		while (x > 0) {
			// System.out.println(x);
			ans += c[x][i];
			x -= lowbit(x);
		}
		return ans;
	}

	static Map<Integer, List<Integer>> map;

	static void mapinit(int chu, int n) {
		map = new HashMap<Integer, List<Integer>>();
		for (int i = chu; i < n + chu; i++) {
			map.put(i, new ArrayList<Integer>());
		}
	}
}

class kattio extends PrintWriter {

	BufferedReader r;
	StringTokenizer st;

	public kattio() {
		this(System.in, System.out);
	}

	public kattio(InputStream i, OutputStream o) {
		super(o);
		r = new BufferedReader(new InputStreamReader(i));
	}

	public kattio(String in, String out) throws IOException {
		super(out);
		r = new BufferedReader(new FileReader(in));
	}

	public String next() {
		try {
			while (st == null || !st.hasMoreTokens()) {
				st = new StringTokenizer(r.readLine());
			}
			return st.nextToken();
		} catch (Exception e) {
			// TODO: handle exception
			return null;
		}
	}

	public int nextint() {
		return Integer.parseInt(next());
	}

	public long nextlong() {
		return Long.parseLong(next());
	}

	public double nextdouble() {
		return Double.parseDouble(next());
	}
}

**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 can cause the code to fail. Here are the steps to create such a test input:

1. **Edge Cases:**
   - Test with the minimum and maximum values for \( n \) and \( k \).
   - Test with all special balls having the same value.
   - Test with all non-special balls having the same value.
   - Test with a mix of values where special and non-special balls have different values.

2. **Scenarios to Maximize Failure Probability:**
   - Test with \( k = 1 \) and \( n \) large.
   - Test with \( k = n \).
   - Test with \( k \) close to \( n/2 \).

Given the constraints and the need for a small-scale test input, we can create a test case that is both simple and challenging. Here is the test input:

```plaintext
1
5 3
10 20 30 40 50
```

**Explanation:**
- **Number of Test Cases:** 1
- **Number of Balls (n):** 5
- **Number of Special Balls (k):** 3
- **Values of Balls:** 10, 20, 30, 40, 50
- **Special Balls:** First 3 balls (10, 20, 30)

This test case is designed to:
- Ensure that the code handles the scenario where the number of special balls is more than half of the total balls.
- Test the code's ability to correctly calculate the expected scores for Alice and Bob when the values are distinct and the game involves multiple turns.

This input should help in exposing any bugs related to the handling of special and non-special balls and the calculation of expected scores.