using namespace std;

#include <set>
#include <map>
#include <list>
#include <deque>
#include <stack>
#include <queue>
#include <cmath>
#include <cctype>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <utility>
#include <iomanip>
#include <fstream>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>

#define oo 1<<30
#define f first
#define s second
#define II inline
#define db double
#define ll long long
#define pb push_back
#define mp make_pair
#define Size(V) ((int)(V.size()))
#define all(v) v.begin() , v.end()
#define CC(v) memset((v),0,sizeof((v)))
#define CP(v,w) memcpy((v),(w),sizeof((w)))
#define FOR(i,a,b) for(int (i)=(a);(i)<=(b);++(i))
#define REP(i, N) for (int (i)=0;(i)<(int)(N);++(i))
#define FORit(it, x) for (__typeof((x).begin()) it = (x).begin(); it != (x).end(); ++it)

#define IN "pedefe.in"
#define OUT "pedefe.out"
#define N_MAX (502)
#define Mod (666013)
#define add(x) ((x) > Mod ? (x) - Mod : (x))

typedef vector<int> VI;
typedef pair<int,int> pi;
typedef vector<string> VS;
template<class T> string toString(T n) {ostringstream ost;ost<<n;ost.flush();return ost.str();}

int X,Y,Z,S1[N_MAX],S2[N_MAX],S3[N_MAX];
int C[3][N_MAX][N_MAX];

II void scan()
{
    freopen(IN,"r",stdin);
    freopen(OUT,"w",stdout);

    scanf("%d%d%d",&X,&Y,&Z);
    FOR(i,1,X) scanf("%d",S1+i);
    FOR(i,1,Y) scanf("%d",S2+i);
    FOR(i,1,Z) scanf("%d",S3+i);
}

II int mod(int x)
{
    for(;x >= Mod;x -= Mod);
    for(;x <= -Mod;x += Mod);
    return x;
}

II void solve()
{
    int A(0),B(1);
    S3[Z+1] = 501;

    FOR(x,1,S3[1])
    FOR(j1,1,X)
    FOR(j2,1,Y)
    {
        if(S1[j1] == x && S1[j1] == S2[j2] && S1[j1] <= S3[1])
            C[2][j1][j2] = C[0][j1-1][j2-1] + 1;
        C[0][j1][j2] = mod(C[2][j1][j2] + C[0][j1-1][j2] + C[0][j1][j2-1] - C[0][j1-1][j2-1]);
    }

    FOR(i,1,Z)
    {
        A ^= 1;
        B ^= 1;

        CC(C[A]);
        for(int j1 = X;j1 >= 1;--j1)
        for(int j2 = Y;j2 >= 1;--j2)
            C[2][j1][j2] = (S1[j1] == S2[j2] && S1[j1] == S3[i]) ? C[B][j1-1][j2-1] + (i == 1) : 0;

        FOR(x,S3[i],S3[i+1])
        {
            bool ok = (x == S3[i] ? true : false);
            FOR(j1,1,X)
            FOR(j2,1,Y)
            {
                if(x != S3[i] && S1[j1] == x && S2[j2] == x)
                    C[2][j1][j2] = add(C[2][j1][j2] + C[A][j1-1][j2-1]),ok = true;
                if(ok)
                    C[A][j1][j2] = mod(C[2][j1][j2] + C[A][j1-1][j2] + C[A][j1][j2-1] - C[A][j1-1][j2-1]);
            }
        }
    }

    C[A][X][Y] += (C[A][X][Y] < 0 ? Mod : 0);
    printf("%d\n",C[A][X][Y]);
}

int main()
{
    scan();
    solve();

    return 0;
}