#include <bits/stdc++.h>
using namespace std;
const int MAXN = 300; // MAXM==MAXN
int n, m;
int a[MAXN + 5];
set<int> D;
// dp[l][r] a[l]~a[r] 最多消除的数量
int dp[MAXN + 5][MAXN + 5];
void work()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    D.clear();
    for (int i = 1; i <= m; i++)
    {
        int d;
        cin >> d;
        D.insert(d);
    }
    // dp[i][i-1] 是长度为 0 的区间
    for (int l = 1; l <= n; l++)
        for (int r = l - 1; r <= n; r++)
            dp[l][r] = 0;
    // 区间dp
    for (int len = 2; len <= n; len++)
        for (int l = 1; l + len - 1 <= n; l++)
        {
            int r = l + len - 1;
            // 直接分两边   l~i i+1~r
            for (int i = l; i <= r - 1; i++)
                dp[l][r] = max(dp[l][r], dp[l][i] + dp[i + 1][r]);
            // 中间消完，头尾消除
            if (dp[l + 1][r - 1] == (r - 1) - (l + 1) + 1 &&
                D.find(a[r] - a[l]) != D.end())
                dp[l][r] = max(dp[l][r], dp[l + 1][r - 1] + 2);
            // 中间挑个位置，消除 a[l],a[i],a[r]
            for (int i = l + 1; i <= r - 1; i++)
            {
                if (a[i] - a[l] == a[r] - a[i] &&
                    D.find(a[i] - a[l]) != D.end() &&
                    dp[l + 1][i - 1] == (i - 1) - (l + 1) + 1 &&
                    dp[i + 1][r - 1] == (r - 1) - (i + 1) + 1)
                    dp[l][r] = max(dp[l][r], dp[l + 1][i - 1] + dp[i + 1][r - 1] + 3);
            }
        }
    cout << dp[1][n] << "\n";
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;
    cin >> T;
    while (T--)
        work();
    return 0;
}