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// Karol Kosinski 2022
#include <bits/stdc++.h>
#define FOR(i,a,b) for(int i=(a),_b=(b);i<_b;++i)
#define FR_(i,a,b) for(int i=(a),_b=(b);i<=_b;++i)
#define FD_(i,b,a) for(int i=(b),_a=(a);i>=_a;--i)
#define ALL(c) (c).begin(),(c).end()
#define SIZE(c) int((c).size())
#define TIE(x...) int x;tie(x)
#define X first
#define Y second
#ifndef ENABLE_DEBUG
#define DEB(k,p,f,x...)
#else
#define DEB(k,p,f,x...) {if(k)printf("--------%4d : %s\n",__LINE__,__FUNCTION__);if(p)f(x);}
#endif
#define DEBL DEB(1,1,void,0)
#define DEBF(f,x...) DEB(1,1,f,x)
#define DEBC(p,x...) DEB(0,p,printf,x)
#define DEBUG(x...) DEB(0,1,printf,x)
using namespace std;
using LL = long long;
using ULL = unsigned long long;
using PII = pair<int, int>;
using TIII = tuple<int, int, int>;

constexpr int NX = 100'005;
constexpr int NX_RED = 1'005;
constexpr int KX = 11;

int A[2][NX], Adj[ 2 * NX ][3];
LL Res[KX];

template <int N>
struct FindUnion
{
    int P[N], S[N];

    void init(int x, int y) { FR_(i,x,y) P[i] = i, S[i] = 1; }
    int find(int x) { return ( x == P[x] ) ? x : P[x] = find( P[x] ); }

    int unite(int x, int y)
    {
        x = find(x), y = find(y);
        if ( S[x] < S[y] ) swap(x, y);
        return S[x] += S[y], P[y] = x;
    }
};

FindUnion< 2 * NX > FU {};

void solve_k1(int n)
{
    
}

void solve_n2(int n2, int k)
{
    FR_(i,1,n2)
    {
        FU.init(i, n2);
        int cnt = 1;
        FR_(j,i+1,n2)
        {
            FOR(x,0,3)
            {
                int u = Adj[j][x];
                if ( i <= u and u < j and FU.find(j) != FU.find(u) )
                {
                    FU.unite( j, u );
                    -- cnt;
                }
            }
            ++ cnt;
            if ( cnt <= k ) ++ Res[cnt];
            DEBUG("[%2d, %2d] : counter = %2d", i, j, cnt);
        }
    }
    Res[1] += n2;
}

int main()
{
    int n, k; scanf("%d%d", &n, &k);
    FOR(i,0,2) FOR(j,0,n)
    {
        int a; scanf("%d", &a);
        A[i][j] = a;
    }
    FOR(i,0,2) FOR(j,0,n)
    {
        int a = A[i][j];
        Adj[a][0] = A[ i ^ 1 ][j];
        Adj[a][1] = A[i][ ( j + 1 ) % n ];
        Adj[a][2] = A[i][ ( j + n - 1 ) % n ];
    }
    // if ( k == 1 )
    // {
        // solve_k1(n);
    // }
    // else
    {
        solve_n2( n * 2, k );
    }
    FR_(i,1,k) printf("%lld ", Res[i]);
    printf("\n");
    return 0;
}