#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef pair<int, int> pi;
#define debug(x) cerr << #x << ": " << x << endl;
#define debug2(x, y) debug(x) debug(y);
#define repn(i, a, b) for(int i = (int)(a); i < (int)(b); i++)
#define rep(i, a) for(int i = 0; i < (int)(a); i++)
#define all(v) v.begin(), v.end()
#define mp make_pair
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define fi first
#define se second
#define sq(x) ((x) * (x))
template<class T> T gcd(T a, T b){ return ((b == 0) ? a : gcd(b, a % b)); }
int dp[90005];
int dp1[1605][1605];
int main(){
ios_base::sync_with_stdio(false);
cin.tie(0);
//freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
int n, m, k;
cin >> n >> m >> k;
vi a(n), b(m);
int sumA = 0;
int mn = 1e9;
rep(i, n) cin >> a[i], sumA += a[i], mn = min(mn, a[i]);
rep(i, m) cin >> b[i];
/*
if(mn < k){
cout << "Impossible" << endl;
return 0;
}
*/
if(n >= 1 && n <= 300 && k <= 300 && m >= 1 && m <= 15){
//subtasks 1 and 2: bitmask
int ans = 1e9;
rep(i, (1 << m)) if(__builtin_popcount(i) >= k){
int sum = 0;
int need = n * k;
int can = 0;
rep(j, m) if(i & (1 << j)){
sum += b[j];
can += min(n, b[j]);
}
if(sum >= sumA && can >= need) ans = min(ans, sum - sumA);
}
if(ans == 1e9) cout << "Impossible" << endl;
else cout << ans << endl;
}
else if(k == 1){
//subtast 3 -> DP
rep(i, 90005) dp[i] = 0;
dp[0] = 1;
rep(i, m){
for(int j = 90005 - b[i]; j >= 0; j--) dp[j + b[i]] |= dp[j];
}
repn(i, sumA, 90005) if(dp[i]){
cout << i - sumA << endl;
return 0;
}
cout << "Impossible" << endl;
}
else{
//subtask 4: DP?
//dp[value][can?]
//value can at most bee 1600, same for can
rep(i, 1605) rep(j, 1605) dp1[i][j] = 0;
dp1[0][0] = 1;
rep(i, m){
//look over all values, add can to them?
int cur = min(n, b[i]);
for(int j = 1605 - b[i]; j >= 0; j--){
rep(k1, 1605) if((k1 + cur) < 1605){
dp1[j + b[i]][k1 + cur] |= dp1[j][k1];
}
}
}
repn(i, sumA, 1605){
repn(j, k * n, 1605) if(dp1[i][j]){
cout << i - sumA << endl;
return 0;
}
}
}
return 0;
}
/*
Things to look out for:
- Integer overflows
- Array bounds
- Special cases
Be careful!
*/
