#include<bits/stdc++.h>
#define ll long long
#define all(a) (a).begin(),(a).end()
using namespace std;

struct BIT {
    int n; vector<int> c;
    BIT(int sz) : n(sz), c(sz + 1) {};
    int lowbit(int x) { return x & -x; }
    void add(int x, int val) {
        while (x <= n) c[x] += val, x += lowbit(x);
    }
    int getsum(int x){
        int ans = 0;
        while (x) ans += c[x], x -= lowbit(x);
        return ans;
    }
    int query(int l, int r){ return getsum(r) - getsum(l - 1); }
    int kth(int k) {//大于等于k的最小位置
    	int res = 0, x = 0; 
    	for (int i = log2(n); i >= 0; --i) {
        	x += 1 << i;                 
        	if (x >= n || res + c[x] >= k) x -= 1 << i;
        	else res += c[x];
    	}
   		return x + 1;
	}
};

void solve()
{
    int n, m;
    cin >> n >> m;
    vector<int> a(n);
    for(int i = 0; i < n; i++) cin >> a[i];

    auto v = a;
    sort(all(v));
    v.erase(unique(all(v)), v.end());
    BIT bit(n + 1);
    int t = 0;
    for(int i = 0; i < n; i++)
    {
        int idx = lower_bound(all(v), a[i]) - v.begin() + 1;
        t += i - bit.getsum(idx);
        bit.add(idx, 1);
    }

    sort(all(a));
    while(m--)
    {
        int k;
        cin >> k;
        vector<int> b(k), cnt1(k), cnt2(k);
        int res = 0;
        for(int i = 0; i < k; i++)
        {
            cin >> b[i];
            cnt1[i] = a.end() - upper_bound(all(a), b[i]);
            cnt2[i] = lower_bound(all(a), b[i]) - a.begin();
            res += cnt1[i];

            // if(k == 3) cerr << i << ' ' << cnt1[i] << ' ' << cnt2[i] << '\n';
        }

        auto v = b;
        sort(all(v));
        v.erase(unique(all(v)), v.end());
        BIT bit(k + 1);
        for(int i = 0; i < k; i++)
        {
            int idx = lower_bound(all(v), b[i]) - v.begin() + 1;
            res += i - bit.getsum(idx);
            bit.add(idx, 1);
        }

        int ans = res;
        // if(k == 3) cerr << "res " << res << '\n';
        for(int i = 0; i < k; i++)
        {
            res -= cnt1[i];
            res += cnt2[i];
            ans = min(ans, res);
            // if(k == 3) cerr << res << '\n';
        }

        cout << ans + t << '\n';
    }
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);cout.tie(nullptr);
    int t = 1;
    //cin >> t;
    while(t--) solve();
    return 0;
}