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 (1<<9)
#define Mod (666013)
#define add_mod(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 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] = (C[2][j1][j2] + C[0][j1-1][j2] + C[0][j1][j2-1] - C[0][j1-1][j2-1]) % Mod;
}
/*
printf("pt %d - final\n",0);
FOR(j1,1,X)
{
FOR(j2,1,Y)
printf("%d ",C[0][j1][j2]);
printf("\n");
} printf("\n");
*/
FOR(i,1,Z)
{
A ^= 1;
B ^= 1;
CC(C[2]);
CC(C[A]);
FOR(j1,1,X)
FOR(j2,1,Y)
if(S1[j1] == S2[j2] && S1[j1] == S3[i])
C[2][j1][j2] = C[B][j1-1][j2-1] + (i == 1);
/*
printf("pt %d - fara suma\n",i);
FOR(j1,1,X)
{
FOR(j2,1,Y)
printf("%d ",C[2][j1][j2]);
printf("\n");
} printf("\n");
*/
FOR(x,S3[i],S3[i+1])
{
FOR(j1,1,X)
FOR(j2,1,Y)
{
if(S1[j1] == x && S1[j1] == S2[j2] && S1[j1] >= S3[i] && S1[j1] <= S3[i+1] && (C[A][j1-1][j2] || C[A][j1][j2-1]) )
C[2][j1][j2] = C[A][j1-1][j2-1] + ( (S1[j1] == S3[i]) ? C[B][j1-1][j2-1] + (i == 1) : 0);
C[A][j1][j2] = (C[2][j1][j2] + C[A][j1-1][j2] + C[A][j1][j2-1] - C[A][j1-1][j2-1]) % Mod;
}
/*
printf("pt %d - final dupa %d\n",i,x);
FOR(j1,1,X)
{
FOR(j2,1,Y)
printf("%d ",C[A][j1][j2]);
printf("\n");
} printf("\n");
*/
}
}
int rez = C[A][X][Y];
printf("%d\n",rez < 0 ? rez + Mod : rez);
}
int main()
{
scan();
solve();
return 0;
}