Piegirl was asked to implement two table join operation for distributed database system, minimizing the network traffic.

Suppose she wants to join two tables, A and B. Each of them has certain number of rows which are distributed on different number of partitions. Table A is distributed on the first cluster consisting of m partitions. Partition with index i has ai rows from A. Similarly, second cluster containing table B has n partitions, i-th one having bi rows from B. 

In one network operation she can copy one row from any partition to any other partition. At the end, for each row from A and each row from B there should be a partition that has both rows. Determine the minimal number of network operations to achieve this.

Input

First line contains two integer numbers, m and n (1 ≤ m, n ≤ 105). Second line contains description of the first cluster with m space separated integers, ai (1 ≤ ai ≤ 109). Similarly, third line describes second cluster with n space separated integers, bi (1 ≤ bi ≤ 109).

Output

Print one integer — minimal number of copy operations.

Examples

Input

2 2
2 6
3 100


Output

11


Input

2 3
10 10
1 1 1


Output

6

Note

In the first example it makes sense to move all the rows to the second partition of the second cluster which is achieved in 2 + 6 + 3 = 11 operations

In the second example Piegirl can copy each row from B to the both partitions of the first cluster which needs 2·3 = 6 copy operations.
#include <bits/stdc++.h>
using namespace std;
const double PI = atan(1) * 4;
const int N = 7e3 + 5;
const double EPS = 1e-9;
int main() {
  int n, m;
  cin >> n >> m;
  vector<long long> A(n), B(m);
  for (int i = 0; i < (int)n; ++i) {
    cin >> A[i];
  }
  for (int i = 0; i < (int)m; ++i) {
    cin >> B[i];
  }
  sort(A.begin(), A.end());
  reverse(A.begin(), A.end());
  sort(B.begin(), B.end());
  reverse(B.begin(), B.end());
  if (A[0] < B[0]) {
    swap(A, B);
    swap(n, m);
  }
  long long sum = 0, ans = 0;
  for (int i = 0; i < (int)m; ++i) {
    sum += B[i];
  }
  ans += sum;
  for (long long i = 1; i <= n - 1; ++i) {
    if (A[i] > sum) {
      ans += sum;
    } else {
      ans += A[i];
    }
  }
  cout << ans << endl;
}
