//思路;把每类出现的次数中相同的出现次数做累加,在用总的数减去小于等于次天数对应的和
//算法:哈希表+前缀和。
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 1e5 + 10;
ll a[N];
void solve(int sum, int q)
{
//哈希表,时间复杂度O1;
unordered_map<int, int> mp;
for (int i = 1; i <= sum; i++)
{
int x;
cin >> x;
assert(x <= 1e6);
mp[x]++;
}
//初始化元素值都为0动态数组。便于后续的前缀和的实现。
vector<int> v(sum + 1, 0);
for (auto&[x, y] :mp)
{
v[y] += y;
}
//前缀和的实现,时间复杂度为O1;
for (int i = 1; i <= sum; i++)
{
v[i] += v[i - 1];
}
int t = sum;
//q次询问
while (q--)
{
int d;
int remain;
cin >> d;
d = min(d, sum);
remain = t - v[d];
cout << remain << ' ';
}
//易犯格式化错误,每轮询问后要进行换行。
cout << endl;
}
int main()
{
//ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int T;
cin>>T;
while(T--)
{
int n, q;
cin >> n >> q;
assert(n <= 2e5);
assert(n <= 2e5);
solve(n, q);
}
return 0;
}