Problem:
Score : 500 points

Problem Statement

On an xy plane, in an area satisfying 0 ≤ x ≤ W, 0 ≤ y ≤ H, there is one house at each and every point where both x and y are integers.

There are unpaved roads between every pair of points for which either the x coordinates are equal and the difference between the y coordinates is 1, or the y coordinates are equal and the difference between the x coordinates is 1.

The cost of paving a road between houses on coordinates (i,j) and (i+1,j) is p_i for any value of j, while the cost of paving a road between houses on coordinates (i,j) and (i,j+1) is q_j for any value of i.

Mr. Takahashi wants to pave some of these roads and be able to travel between any two houses on paved roads only. Find the solution with the minimum total cost.

Constraints

1 ≦ W,H ≦ 10^5

1 ≦ p_i ≦ 10^8(0 ≦ i ≦ W-1)

1 ≦ q_j ≦ 10^8(0 ≦ j ≦ H-1)

p_i (0 ≦ i ≦ W−1) is an integer.

q_j (0 ≦ j ≦ H−1) is an integer.

Input

Inputs are provided from Standard Input in the following form.

W H
p_0
:
p_{W-1}
q_0
:
q_{H-1}

Output

Output an integer representing the minimum total cost.

Sample Input 1

2 2
3
5
2
7

Sample Output 1

29

It is enough to pave the following eight roads.

Road connecting houses at (0,0) and (0,1)

Road connecting houses at (0,1) and (1,1)

Road connecting houses at (0,2) and (1,2)

Road connecting houses at (1,0) and (1,1)

Road connecting houses at (1,0) and (2,0)

Road connecting houses at (1,1) and (1,2)

Road connecting houses at (1,2) and (2,2)

Road connecting houses at (2,0) and (2,1)

Sample Input 2

4 3
2
4
8
1
2
9
3

Sample Output 2

60

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

using VS = vector<string>;    using LL = long long;
using VI = vector<int>;       using VVI = vector<VI>;
using PII = pair<int, int>;   using PLL = pair<LL, LL>;
using VL = vector<LL>;        using VVL = vector<VL>;

#define ALL(a)  begin((a)),end((a))
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define SZ(a) int((a).size())
#define SORT(c) sort(ALL((c)))
#define RSORT(c) sort(RALL((c)))
#define UNIQ(c) (c).erase(unique(ALL((c))), end((c)))
#define FOR(i, s, e) for (int(i) = (s); (i) < (e); (i)++)
#define FORR(i, s, e) for (int(i) = (s); (i) > (e); (i)--)
#define debug(x) cerr << #x << ": " << x << endl
const int INF = 1e9;                          const LL LINF = 1e16;
const LL MOD = 1000000007;                    const double PI = acos(-1.0);
int DX[8] = { 0, 0, 1, -1, 1, 1, -1, -1 };    int DY[8] = { 1, -1, 0, 0, 1, -1, 1, -1 };

/* -----  2018/03/03  Problem: code-festival-2016-qual-b_c / Link: https://code-festival-2016-qualb.contest.atcoder.jp/tasks/codefestival_2016_qualB_c?lang=en  ----- */
/* ------問題------

xy平面上の0≦x≦W,0≦y≦Hをみたす領域にあるx,yともに整数である点すべてに、ひとつずつ家があります。
x座標が等しくy座標の差が1であるか、y座標が等しくx座標の差が1であるような2点の組のうち、両方の点に家が存在するような全てのものに対し、その2点の間には舗装されていない道路があります。
座標(i,j)と(i+1,j)にある家の間の道路を舗装するのにはjの値にかかわらずコストがpi、 座標(i,j)と(i,j+1)にある家の間の道路を舗装するのにはiの値にかかわらずコストがqjかかります。
高橋君は、このうちいくつかの道路を舗装し、舗装された道路のみを通って任意の2つの家の間を行き来できるようにしたいです。 かかるコストの総和の最小値を求めてください。

-----問題ここまで----- */
/* -----解説等-----



----解説ここまで---- */

LL W, H;

LL ans = 0LL;

int main() {
	cin.tie(0);
	ios_base::sync_with_stdio(false);

	cin >> W >> H;
	vector<PLL> a;
	FOR(i, 0, W) {
		int v; cin >> v;
		a.push_back(PLL(v, 0));
	}
	FOR(i, 0, H) {
		int v; cin >> v;
		a.push_back(PLL(v, 1));
	}

	SORT(a);
	H++, W++;

	FOR(i, 0, SZ(a)) {
		if (a[i].second == 0) {
			ans += H*a[i].first;
			W--;
		}
		else {
			ans += W*a[i].first;
			H--;
		}
	}

	cout << ans << "\n";

	return 0;
}