#include <string>
#include <vector>
#include <algorithm>
#include <numeric>
#include <set>
#include <map>
#include <queue>
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <cctype>
#include <cassert>
#include <limits>
#include <functional>
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))
#define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))
#if defined(_MSC_VER) || __cplusplus > 199711L
#define aut(r,v) auto r = (v)
#else
#define aut(r,v) __typeof(v) r = (v)
#endif
#define each(it,o) for(aut(it, (o).begin()); it != (o).end(); ++ it)
#define all(o) (o).begin(), (o).end()
#define pb(x) push_back(x)
#define mp(x,y) make_pair((x),(y))
#define mset(m,v) memset(m,v,sizeof(m))
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3fLL
using namespace std;
typedef vector<int> vi; typedef pair<int, int> pii; typedef vector<pair<int, int> > vpii; typedef long long ll;
template<typename T, typename U> inline void amin(T &x, U y) { if(y < x) x = y; }
template<typename T, typename U> inline void amax(T &x, U y) { if(x < y) x = y; }

int main() {
    typedef long double Num;
    vector<Num> log2s(2001);
    rer(n, 0, 2000)
        log2s[n] = log2(Num(n));
    int T;
    scanf("%d", &T);
    for(int ii = 0; ii < T; ++ ii) {
        int N; int M;
        scanf("%d%d", &N, &M);
        vector<bool> bomb(N);
        rep(i, M) {
            int X;
            scanf("%d", &X);
            bomb[X] = true;
        }
        vector<int> nextPos(N + 1);
        nextPos[N] = N;
        for(int i = N - 1; i >= 0; -- i) {
            nextPos[i] = bomb[i] ? i : nextPos[i + 1];
        }
        vector<Num> dp(N + 1, -1e99);
        dp[0] = 0;
        rep(i, N) {
            Num x = dp[i];
            if(x == -1e99) continue;
            int pos = nextPos[i];
            if(pos == N)
                continue;
            int R = nextPos[pos + 1];
            if(R == N) {
                amax(dp[N], x + log2s[N - i]);
                continue;
            }
            amax(dp[pos + 1], x + log2s[pos + 1 - i]);
            for(int j = pos + 1; j < R; ++ j)
                amax(dp[j + 1], x + log2s[j + 1 - i]);
        }
        cout << (long long)floor(dp[N] * 1000000) << '\n';
    }
    return 0;
}