#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
//9 pts: 1 <= M <= 2
/*
int main()
{
int N, M, K;
cin >> N >> M >> K;
int A[N], B[M];
int Asum = 0;
bool any1 = 0;
for(int i = 0; i < N; i++)
{
cin >> A[i];
Asum += A[i];
if(A[i] == 1) any1 = 1;
}
for(int j = 0; j < M; j++) cin >> B[j];
if(M == 1)
{
if(K == 1 && B[0] >= Asum) cout << B[0] - Asum << '\n';
else cout << "Impossible\n";
}
else if(M == 2)
{
if(K == 1)
{
if(B[0] > B[1]) swap(B[0], B[1]);
if(Asum <= B[0]) cout << B[0] - Asum << '\n';
else if(Asum <= B[1]) cout << B[1] - Asum << '\n';
else if(Asum <= B[0] + B[1]) cout << B[0] + B[1] - Asum << '\n';
else cout << "Impossible\n";
}
else if(K == 2)
{
if(!any1 && B[0] >= N && B[1] >= N && B[0] + B[1] >= Asum) cout << B[0] + B[1] - Asum << '\n';
else cout << "Impossible\n";
}
else cout << "Impossible\n";
}
}
*/
//20 pts: K = 1
/*
int main()
{
int N, M, K;
cin >> N >> M >> K;
int A[N], B[M];
int Asum = 0;
for(int i = 0; i < N; i++)
{
cin >> A[i];
Asum += A[i];
}
for(int j = 0; j < M; j++) cin >> B[j];
vector<int> possible(90001, 0);
possible[0] = 1;
int res = 90000;
for(int j = 0; j < M; j++)
{
for(int i = 90000; i >= B[j]; i--)
{
if(!possible[i - B[j]]) continue;
possible[i] = 1;
if(i >= Asum) res = min(res, i - Asum);
}
}
if(res == 90000) cout << "Impossible\n";
else cout << res << '\n';
}
*/
/*
Condition for value K, meals 1, 2, ... N and chefs* 1, 2, ... y to be compatible (* - chefs are re-numbered)
A[i] >= K
sum{j:chefs | max(B[j], N)} >= N*K
sum{j:chefs | B[j]} >= sum{i:meals | A[i]}
To minimize: sum{j:chefs | B[j]} - sum{i:meals | A[i]
*/
//100 pts
int main()
{
int N, M, K;
cin >> N >> M >> K;
int A[N], B[M];
bool impossible = 0;
int Asum = 0;
for(int i = 0; i < N; i++)
{
cin >> A[i];
Asum += A[i];
if(A[i] < K) impossible = 1;
}
for(int j = 0; j < M; j++) cin >> B[j];
if(impossible)
{
cout << "Impossible\n";
return 0;
}
//vector<bool> chef_sum(90001, 0);
//chef_sum[0] = 1;
vector<int> variety_sum(90001, -1); //sum over all dishes for number of different chefs for given total chef time sum
variety_sum[0] = 0;
for(int j = 0; j < M; j++)
{
for(int i = 90000; i >= B[j]; i--)
{
if(variety_sum[i - B[j]] >= 0)
{
variety_sum[i] = max(variety_sum[i], variety_sum[i - B[j]] + min(B[j], N));
}
}
}
for(int i = Asum; i <= 90000; i++)
{
if(variety_sum[i] >= N*K)
{
cout << i - Asum << '\n';
return 0;
}
}
cout << "Impossible\n";
}
