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;
int a[100005], b[100005], tmp[100005];
int main() {
  int n, m, i;
  while (~scanf("%d%d", &n, &m)) {
    long long sa = 0, sb = 0;
    for (i = 0; i < n; i++) scanf("%d", &a[i]), sa += a[i];
    for (i = 0; i < m; i++) scanf("%d", &b[i]), sb += b[i];
    sort(a, a + n, greater<int>());
    sort(b, b + m, greater<int>());
    if (a[0] < b[0])
      memcpy(tmp, a, sizeof(a)), memcpy(a, b, sizeof(b)),
          memcpy(b, tmp, sizeof(tmp)), swap(sa, sb), swap(n, m);
    long long ans = sb;
    for (i = 1; i < n; i++) ans += a[i] > sb ? sb : a[i];
    cout << ans << endl;
  }
  return 0;
}
