#include <bits/stdc++.h>
#include <atcoder/modint>
using namespace std;
using namespace atcoder;
using ll = long long;
using mint = modint998244353;
#define rep(i,n) for (int i = 0; i < n; i++)
#define rrep(i,n) for (int i = n-1; i >= 0; i--)
#define rep2(i,a,b) for (int i = a; i < b; i++)
#define rrep2(i,a,b) for (int i = a-1; i >= b; i--)
#define rep3(i,a,b,c) for (int i = a; i < b; i+=c)
#define rrep3(i,a,b,c) for (int i = a-1; i >= b; i-=c)
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()
template<class T> bool chmax(T &a, T b){if (a < b){a = b;return true;} else return false;}
template<class T> bool chmin(T &a, T b){if (a > b){a = b;return true;} else return false;}

int n,x,y,A[1<<18],B[1<<18],C[20],D[20];
mint dp[200010][20][2];
int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cin>>n>>x>>y;
    rep(i,x){
        cin>>A[i];
        rep(j,20){
            if(A[i]>>j&1)C[j]++;
        }
    }
    rep(i,y){
        cin>>B[i];
        rep(j,20){
            if(B[i]>>j&1)D[j]++;
        }
    }
    rep(i,20)dp[0][i][0]=1;
    rep(i,n){
        rep(j,20){
            dp[i+1][j][1]=x*dp[i][j][1]*D[j]+C[j]*dp[i][j][0]*D[j];
            dp[i+1][j][0]=x*(dp[i][j][1]+dp[i][j][0])*(y-D[j])+(x-C[j])*dp[i][j][0]*D[j];
        }
    }
    mint ans=0;
    rep(i,20)ans+=(1<<i)*dp[n][i][1];
    cout<<ans.val()<<"\n";
}