#include <bits/stdc++.h>
using namespace std;
const int N = 2e5;
int a[N], b[N];
long long k, l, ans, sum, k1, l1;
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    k += a[i];
  }
  for (int i = 0; i < m; i++) {
    scanf("%d", &b[i]);
    l += b[i];
  }
  sort(a, a + n);
  sort(b, b + m);
  int i = m - 1;
  l1 = l;
  for (i = m - 1; i >= 0 && b[i] > k; i--) l1 -= b[i];
  if (i < m - 1)
    i++;
  else
    l1 -= b[m - 1];
  sum = (long long)(m - i) * k + l1;
  k1 = k;
  for (i = n - 1; i >= 0 && a[i] > l; i--) k1 -= a[i];
  if (i < n - 1)
    i++;
  else
    k1 -= a[n - 1];
  ans = (n - i) * l + k1;
  printf("%I64d", min(ans, sum));
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5;
int a[N], b[N];
long long k, l, ans, sum, k1, l1;
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    k += a[i];
  }
  for (int i = 0; i < m; i++) {
    scanf("%d", &b[i]);
    l += b[i];
  }
  sort(a, a + n);
  sort(b, b + m);
  int i = m - 1;
  l1 = l;
  for (i = m - 1; i >= 0 && b[i] > k; i--) l1 -= b[i];
  if (i < m - 1)
    i++;
  else
    l1 -= b[m - 1];
  sum = (long long)(m - i) * k + l1;
  k1 = k;
  for (i = n - 1; i >= 0 && a[i] > l; i--) k1 -= a[i];
  if (i < n - 1)
    i++;
  else
    k1 -= a[n - 1];
  ans = (n - i) * l + k1;
  printf("%I64d", min(ans, sum));
  return 0;
}

```