I - GCD Expectation ZOJ - 3868

Edward has a set of n integers {a₁, a₂,...,a_n}. He randomly picks a nonempty subset {x₁, x₂,…,x_m} (each nonempty subset has equal probability to be picked), and would like to know the expectation of [gcd(x₁, x₂,…,x_m)]^k.

Note that gcd(x₁, x₂,…,x_m) is the greatest common divisor of {x₁, x₂,…,x_m}.

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains two integers n, k (1 ≤ n, k ≤ 10⁶). The second line contains n integers a₁, a₂,…,a_n (1 ≤ a_i ≤ 10⁶).

The sum of values max{a_i} for all the test cases does not exceed 2000000.

Output

For each case, if the expectation is E, output a single integer denotes E · (2ⁿ - 1) modulo 998244353.

Sample Input

1
5 1
1 2 3 4 5

Sample Output

42

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn=1e6+10;
const int inf=0x3f3f3f3f;
const int mod=998244353;
ll num_mul[maxn],sum[maxn];
int val[maxn];
/*
The quantity must also be divided at the same time, because it may cause the value to become very "large" later, resulting in no way to calculate
Using the principle of inclusion and exclusion, it feels more like screening, i remove the multiple of i
*/

void init(){
	memset(num_mul,0,sizeof(num_mul));
	memset(sum,0,sizeof(sum));
}

ll q_pow(ll base,int n){
	ll ans=1;
	while(n){
		if(n&1){
			ans=(ans*base)%mod;
		}
		base=(base*base)%mod;
		n>>=1;
	}
	return ans;
}

int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		init();
		int n,k,mv=0;
		scanf("%d %d",&n,&k);
		for(int i=1;i<=n;i++){
			scanf("%d",&val[i]);
			sum[val[i]]++;
			mv=max(mv,val[i]);
		}
		ll ans=0;
		for(int i=mv;i>=1;i--){
			int cnt=0;
			for(int j=i;j<=mv;j+=i){
				cnt+=sum[j];
				num_mul[i]=(num_mul[i]-num_mul[j]+mod)%mod;;
			}
			num_mul[i]=(num_mul[i]+q_pow((ll)2,cnt)-1)%mod;
			ans=(ans+num_mul[i]*q_pow((ll)i,k))%mod;
		}
		printf("%lld\n",ans);
	}	
	return 0;
}