两句话题意 感觉这一题的套路很强

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;


#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
LL quick_pow(LL a, LL b, LL MOD) {
    LL ans = 1;
    LL base = a;
    while (b) {
        if (b & 1) {
            ans = ans * base % MOD;
        }
        b >>= 1;
        base = base * base % MOD;
    }
    return ans;
}
const int maxn = 2e6 + 20;
int ans[maxn], num[maxn];
const int MOD = 10000007;
void work() {
    int n, k;
    scanf("%d%d", &n, &k);
    int mx = -inf;
    for (int i = 1; i <= n; ++i) {
        int x;
        scanf("%d", &x);
        num[x]++;
        mx = max(mx, x);
    }
    LL res = 0;
    if (num[0]) {
        ans[0] = (quick_pow(2, num[0], MOD) - 1 + MOD) % MOD;
        res = ans[0];
    }
    for (int i = mx; i >= 1; --i) {
        int has = num[0];
        for (int j = i; j <= mx; j += i) has += num[j];
        ans[i] = (quick_pow(2, has, MOD) - 1 + MOD) % MOD;
        for (int j = 2 * i; j <= mx; j += i) {
            ans[i] = (ans[i] - ans[j] + MOD) % MOD;
        }
        ans[i] = (ans[i] - ans[0] + MOD) % MOD;
        res = (res + quick_pow(i, k, MOD) * ans[i] % MOD) % MOD;
    }
    cout << res << endl;
    for (int i = 0; i <= mx; ++i) ans[i] = num[i] = 0;
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    int t;
    scanf("%d", &t);
    while (t--) {
        work();
//        printf("\n");
    }
    return 0;
}