Let‘sPlayCurling

$$\begin{gathered} 对于公式|c-a_i|<|c-b_j|我们可以用语言描述 \\ 即a_i是离c最近的点，和b_j比较起来 \\ 对于两个红球我们将c放在中间，两个红球都能计数 \\ 对于一个篮球一个红球我们要尽量靠近红球才能算如贡献 \\ 对于两个蓝球，这个时候不产生任何贡献，因为没有比这两个蓝球离根据更近的了 \\ 这样答案问题就转化为求最长的红球个数 \\ 我的问题在于一直没有转化描述思考的过程非常艰难 \end{gathered}$$

这种绝对值的状态用距离来代替

// Problem: Let'sPlayCurling
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/22904/1019
// Memory Limit: 524288 MB
// Time Limit: 2000 ms
// 2022-06-28 12:36:41
// 
// Powered by CP Editor (https://cpeditor.org)

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

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;   
typedef vector<ll> Vll;               
typedef vector<pair<int, int> > vpii;
typedef vector<pair<ll, ll> > vpll;                        

const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const double pi  = acos(-1.0);
inline ll ksc(ll x,ll y,ll mod)
{
    ll ans = 0;
    while (y) {
    	if (y & 1)
    		ans = (ans + x) %mod;
    	y >>= 1;
    	x = (x + x) %mod;
	}
	return ans;
}
inline ll qmi (ll a, ll b) {
	ll ans = 1;
	while (b) {
		if (b & 1) ans = ans * a;
		a = a * a;
		b >>= 1;
	}
	return ans;
}
inline int read () {
	int x = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
	while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
	return f?-x:x;
}
template<typename T> void print(T x) {
	if (x < 0) putchar('-'), x = -x;
	if (x >= 10) print(x/10);
	putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
	return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
	return (a + b) %mod;
}
// inline ll inv (ll a) {
	// return qmi(a, mod - 2);
// }
int a[100005];
int b[100005];
int n, m;

void solve() {
	cin >> n >> m;
	map<int, int> p;
	for (int i = 1; i <= n; i ++)
	{
		int x;
		cin >> x;
		p[x] ++;
	}
	for (int i = 1; i <= m; i ++)
	{
		int x;
		cin >> x;
		p[x] = 0;
	}
	int ans = 0;
	int cnt = 0;
	for (auto [x, st]: p) {
		if (st == 0) cnt = 0;
		else {
			cnt += st;
		}
		ans = max (ans, cnt);
	}
	if (ans == 0) {
		puts("Impossible");
		return;
	}
	cout << ans << endl;
}
int main () {
	// ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);
    int t;
    t =1;
    cin >> t;
    while (t --) solve();
    return 0;
}