#include <bits/stdc++.h>
using namespace std;
#define repd(i,a,b) for (ll i=(a);i<(b);i++)
#define rep(i,n) repd(i,0,n)
#define all(x) (x).begin(),(x).end()
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
typedef long long ll;
typedef pair<ll,ll> P;
typedef vector<ll> vec;
using Graph = vector<vector<ll>>;
const long long INF = 1LL<<60;
const long long MOD = 1000000007;

// auto mod int
// https://youtu.be/L8grWxBlIZ4?t=9858
// https://youtu.be/ERZuLAxZffQ?t=4807 : optimize
// https://youtu.be/8uowVvQ_-Mo?t=1329 : division
const int mod = 998244353;
struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
  mint operator+(const mint a) const { return mint(*this) += a;}
  mint operator-(const mint a) const { return mint(*this) -= a;}
  mint operator*(const mint a) const { return mint(*this) *= a;}
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const { return pow(mod-2);}
  mint& operator/=(const mint a) { return *this *= a.inv();}
  mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}

int nxt[2005][2005];
mint dp[2005][2005];


int main()
{  
    ios::sync_with_stdio(false);
    cin.tie(0);
    ll n,m,k;
    cin>>n>>m>>k;
    vec s(n);
    rep(i,n){
      cin>>s[i];
      s[i]--;
    }
    vec a(m),b(m);
    rep(i,m)cin>>a[i];
    rep(i,m)cin>>b[i];
    for(ll i=n;i>=1;i--){
      for(ll j=i-1;j>=0;j--){
        nxt[j][s[i-1]]=i;
        if(j==0)continue;
        if(s[i-1]==s[j-1])break;
      }
    }
    dp[0][0]=1;
    rep(i,n){
      rep(j,m){
        if(nxt[i][j]==0)continue;
        rep(l,k+1){
          if(l+a[j]>k&&l+b[j]>k)break;
          if(l+a[j]<=k)dp[nxt[i][j]][l+a[j]]+=dp[i][l];
          if(l+b[j]<=k)dp[nxt[i][j]][l+b[j]]+=dp[i][l];
        }
      }
    }
    mint ans=0;
    rep(i,2005)ans+=dp[i][k];
    cout<<ans<<endl;
    return 0;
}