#include <bits/stdc++.h>
using namespace std;
#define int long long
#define vo vector
#define pb push_back
#define se second
#define fi first
#define all(v) v.begin(), v.end()
typedef vector<int> vi;
typedef pair<int, int> pii;
#define rep(i, a, b) for(int i=(a); i<b; i++)
#define pr1(x) cerr << #x << '=' << x << ' ';
//for google contests
#define umap unordered_map
#define uset unordered_set
#define repd(i, a, b) for(int i=(b-1); i >= a; i--)
void _pr(signed x) {cerr << x;}
void _pr(long long x) {cerr << x;}
void _pr(size_t x) {cerr << x;}
void _pr(double x) {cerr << x;}
void _pr(char x) {cerr << '\'' << x << '\'';}
void _pr(const char* x) {cerr << x;}
void _pr(bool x) {cerr << (x ? "true" : "false");}
template<typename T, typename V> void _pr(const pair<T, V> &x);
template<typename T, typename V> void _pr(const pair<T, V> &x) {cerr << "\e[95m" << "[ "; _pr(x.first); cerr << ", "; _pr(x.second); cerr << "\e[95m" << ']';}
template<typename T> void _pr(const T &x) {int F=0; cerr << '{'; for(auto &i: x) cerr << (F++ ? ", " : ""), _pr(i); cerr << "\e[91m" << '}';}
template <typename T, typename... V> void _pr(T t, V... v) {_pr(t); if(sizeof...(v)) cerr << ", "; _pr(v...);}
#define pr(x...) cerr << "\e[91m" << __func__ << ':' << __LINE__ << " [" << #x << "] = ["; _pr(x); cerr << "\e[91m" << ']' << "\033[0m" << endl;
mt19937 mt_rng(chrono::steady_clock::now().time_since_epoch().count());
int rand(int l, int r){return uniform_int_distribution<int>(l, r)(mt_rng);}
//go for outline with ;, then details
int const inf = LLONG_MAX, mxn = 300;
int n, m, k, meal[mxn], mxhours[mxn], sum, ans = inf;
void fulsol(){
int mxval = 90310; vo<pii> dp(mxval); dp[0] = {1, 0};
rep(chef, 0, m){
repd(i, 0, mxval){
if(dp[i].fi){
int nw = mxhours[chef]+i, uniqhours = dp[i].se+min(n, mxhours[chef]);
if(nw < mxval){
dp[nw].fi = 1;
dp[nw].se = max(dp[nw].se, uniqhours);
}
}
}
}
rep(i, sum, mxval)
if(dp[i].fi && dp[i].se>=k*n){
ans = i-sum; break;
}
}
signed main(){
cin.tie(0)->sync_with_stdio(0);
cin>>n>>m>>k;
rep(i, 0, n) {
cin>>meal[i]; sum+=meal[i];
}
rep(i, 0, m) cin>>mxhours[i];
rep(i, 0, n){
if(meal[i] < k){
cout << "Impossible"; exit(0);
}
}
fulsol();
if(ans == inf) cout << "Impossible";
else cout << ans;
}
/*
subtask 1: case-analysis for m,k in {1, 2}
subtask 2: subset bruteforce, use bitmask. see that we can also get prev subtask if we create a function
f(vi chefs, sumchefhours) that returns wasted time
subtask 3: subset sum and find biggest r=subsetsum such that r<=sumhours. time complexity: maxval*ppl = (maxA*n)*m <= 3e7
subtask 4: dp[sum] = {a1, a2, ... , an} where a1 is how many chefs working on person 1. time: mxsum * chefs * dishes = 40^4
subtask 5: dp[sum] = {x, y};
x = if it is reachable
y = maximum sum of hours we can get such that no person gets the same chef twice = sum of min(n, mxhours[i])} for all chefs i included
*/
