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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define _GLIBCXX_DEBUG
using namespace std;
using ll = long long;
using vec = vector<ll>;
using vect = vector<double>;
using Graph = vector<vector<ll>>;
#define loop(i, n) for (ll i = 0; i < n; i++)
#define Loop(i, m, n) for (ll i = m; i < n; i++)
#define pool(i, n) for (ll i = n; i >= 0; i--)
#define Pool(i, m, n) for (ll i = n; i >= m; i--)
#define mod 1000000007ll
//#define mod 998244353ll
#define flagcount(bit) __builtin_popcount(bit)
#define flag(x) (1ll << x)
#define flagadd(bit, x) bit |= flag(x)
#define flagpop(bit, x) bit &= ~flag(x)
#define flagon(bit, i) bit &flag(i)
#define flagoff(bit, i) !(bit & (1ll << i))
#define all(v) v.begin(), v.end()
#define low2way(v, x) lower_bound(all(v), x)
#define high2way(v, x) upper_bound(all(v), x)
#define idx_lower(v, x) (distance(v.begin(), low2way(v, x)))  //配列vでx未満の要素数を返す
#define idx_upper(v, x) (distance(v.begin(), high2way(v, x))) //配列vでx以下の要素数を返す
#define idx_lower2(v, x) (v.size() - idx_lower(v, x))         //配列vでx以上の要素数を返す
#define idx_upper2(v, x) (v.size() - idx_upper(v, x))         //配列vでxより大きい要素の数を返す
#define putout(a) cout << a << endl
#define Sum(v) accumulate(all(v), 0ll)
ll ctoi(char c)
{
    if (c >= '0' && c <= '9')
    {
        return c - '0';
    }
    return -1;
}
template <typename T>
string make_string(T N)
{
    string ret;
    T now = N;
    while (now > 0)
    {
        T x = now % 10;
        ret += (char)('0' + x);
        now /= 10;
    }
    reverse(all(ret));
    return ret;
}
template <typename T>
T gcd(T a, T b)
{
    if (a % b == 0)
    {
        return (b);
    }
    else
    {
        return (gcd(b, a % b));
    }
}
template <typename T>
T lcm(T x, T y)
{
    T z = gcd(x, y);
    return x * y / z;
}
template <typename T>
bool primejudge(T n)
{
    if (n < 2)
        return false;
    else if (n == 2)
        return true;
    else if (n % 2 == 0)
        return false;
    double sqrtn = sqrt(n);
    for (T i = 3; i < sqrtn + 1; i++)
    {
        if (n % i == 0)
        {
            return false;
        }
        i++;
    }
    return true;
}
template <typename T>
bool chmax(T &a, const T &b)
{
    if (a < b)
    {
        a = b; // aをbで更新
        return true;
    }
    return false;
}
template <typename T>
bool chmin(T &a, const T &b)
{
    if (a > b)
    {
        a = b; // aをbで更新
        return true;
    }
    return false;
}
//場合によって使い分ける
//const ll dx[4]={1,0,-1,0};
//const ll dy[4]={0,1,0,-1};
const ll dx[8] = {1, 1, 0, -1, -1, -1, 0, 1};
const ll dy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
//2次元配列の宣言
//vector<vector<ll>> field(h, vector<ll>(w));

int main()
{
    cout << fixed << setprecision(30);
    ll W, H;
    cin >> W >> H;
    vector<pair<ll, bool>> pairs(W + H);
    vector<ll> p(W), q(H);
    loop(i, W)
    {
        cin >> p[i];
        pairs[i] = make_pair(p[i], 1);
    }
    loop(i, H)
    {
        cin >> q[i];
        pairs[W + i] = make_pair(q[i], 0);
    }
    sort(all(pairs));
    ll a = W + 1, b = H + 1;
    ll ans = 0;
    for (ll i = 0; a && b; i++)
    {
        auto p = pairs[i];
        if (p.second)
        {
            ans += b * p.first;
            a--;
        }
        else
        {
            ans += a * p.first;
            b--;
        }
    }
    putout(ans);
    return 0;
}