//Author: Oliver-Matis Lill
//O(N M B_max) DP solution. Finds maximum "diversity box filled" for a given total hour count
#include<iostream>
#include<vector>
#include<numeric>
#include<algorithm>
using namespace std;
const int INF = 1000000000;
int sum(vector<int> &a) {
return accumulate(a.begin(), a.end(), 0);
}
int min(vector<int> &a) {
return *min_element(a.begin(), a.end());
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int n, m, k;
cin >> n >> m >> k;
vector<int> A(n), B(m);
for(int i=0; i<n; i++) {
cin >> A[i];
}
for(int i=0; i<m; i++) {
cin >> B[i];
}
if(min(A) < k) {
cout << "Impossible\n";
return 0;
}
vector<int> dp((m+1)*301, -1);
dp[0] = 0;
for(int i=0; i<m; i++) {
for(int j=dp.size()-B[i]-1; j >= 0; j--) {
if(dp[j] != -1) {
dp[j+B[i]] = max(dp[j+B[i]], dp[j] + min(B[i], n));
}
}
}
int result = INF;
int a_sum = sum(A);
for(int j=a_sum; j<(int)dp.size(); j++) {
if(dp[j] >= n*k) {
result = min(result, j - a_sum);
}
}
if(result == INF) {
cout << "Impossible\n";
} else {
cout << result << '\n';
}
return 0;
}
