#include<bits/stdc++.h>
//#define ll int 
#define inf 0x3f3f3f3f
#define endl '\n'
#define ull unsigned long long 
#define pll pair<ll ,ll>
#define pii pair<int, int>
#define lazy ios::sync_with_stdio(0);cin.tie(0); cout.tie(0);
using namespace std;

typedef long long ll;
const ll N = 1e5 + 7;
const ll M = 2e3 + 7;
const double eps = 1e-6;
const ll mod = 998244353;
ll _ = 1, n, m, a[N], tot, num[N], cntl[2 * N], cntr[2 * N], suml[2 * N], sumr[2 * N];
vector<ll> v[N];
map<ll, ll> mp;

template<class T>
struct BIT {
    ll len;
    vector<T> c;
    BIT(int x){
        len=x;
        c.resize(len + 1);
    }
    void init(ll x){
        len = x;
        c.clear();
        c.resize(len + 1);
    }// 重构数组
    ll lowbit(ll x){
        return x & (-x);
    }
    void modify(ll x, ll y){// 单点加
        assert(x <= len);
        while (x <= len){
            c[x] += y;
            x += lowbit(x);
        }
    }
    T query(ll x){// 1-x前缀和
        assert(x >= 0);
        if (x == 0)return 0;
        T res = 0;
        while (x){
            res += c[x];
            x -= lowbit(x);
        }
        return res;
    }
    T query(ll l, ll r){return query(r) - query(l - 1);}
};


void doit(ll x){
    if (mp.find(x) == mp.end())mp[x] = ++tot;
}


void solve() {
    cin >> m >> n;
    vector<ll> tempv;
    for (int i = 1; i <= m; i++)cin >> a[i], tempv.push_back(a[i]);
    for (int i = 1; i <= n; i++){
        cin >> num[i];
        v[i].push_back(0);
        for (int j = 1, x; j <= num[i]; j++){
            cin >> x;
            tempv.push_back(x);
            v[i].push_back(x);
        }
    }
    sort(tempv.begin(), tempv.end());
    for (auto i : tempv)doit(i);
    for (int i = 1; i <= m; i++)a[i] = mp[a[i]];
    for (int i = 1; i <= n; i++){
        for (int j = 1; j <= num[i]; j++){
            v[i][j] = mp[v[i][j]];
        }
    }
    BIT<ll> b(tot), bl(tot), br(tot);
    ll sum = 0;
    for (int i = m; i; i--){
        sum += b.query(a[i] - 1);
        b.modify(a[i], 1);
        bl.modify(a[i], 1);// 把数放在b数组的左边的逆序对个数
        br.modify(tot + 1 - a[i], 1);// 把数放在b数组的右边的逆序对个数
    }
    for (int i = 1; i <= n; i++){
        ll cnt = 0;
        BIT<ll> bn(tot);
        for (int j = num[i]; j; j--){
            cnt += bn.query(v[i][j] - 1);
            bn.modify(v[i][j], 1);
        }
        for (int j = 0; j <= num[i] + 1; j++)cntl[j] = cntr[j] = suml[j] = sumr[j] = 0;
        for(int j = 1; j <= num[i]; j++){
            cntl[j] = bl.query(v[i][j] - 1);
            suml[j] = suml[j - 1] + cntl[j];
            // cerr << j << " cntl " << cntl[j] << " suml " << suml[j] << endl; 
        }
        for (int j = num[i]; j; j--){
            cntr[j] = br.query(tot + 1 - v[i][j] - 1);
            sumr[j] = sumr[j + 1] + cntr[j];
            // cerr << j << " cntr " << cntr[j] << " sumr " << sumr[j] << endl; 
        }
        ll ans = 1e18;
        for (int j = 0; j <= num[i]; j++){
            ll all = sum + cnt;
            all += suml[j];
            all += sumr[j + 1];
            // cerr << j << " " << all << endl;
            ans = min(ans, all);
        }
        cout << ans << endl;
    }
}



int main() {
    lazy;
    // cin >> _;
    while (_--)solve();
}