There are no heroes in this problem. I guess we should have named it "To Zero".

You are given two arrays $a$ and $b$, each of these arrays contains $n$ non-negative integers.

Let $c$ be a matrix of size $n \times n$ such that $c_{i,j} = |a_i - b_j|$ for every $i \in [1, n]$ and every $j \in [1, n]$.

Your goal is to transform the matrix $c$ so that it becomes the zero matrix, i. e. a matrix where every element is exactly $0$. In order to do so, you may perform the following operations any number of times, in any order:

choose an integer $i$, then decrease $c_{i,j}$ by $1$ for every $j \in [1, n]$ (i. e. decrease all elements in the $i$-th row by $1$). In order to perform this operation, you pay $1$ coin;

choose an integer $j$, then decrease $c_{i,j}$ by $1$ for every $i \in [1, n]$ (i. e. decrease all elements in the $j$-th column by $1$). In order to perform this operation, you pay $1$ coin;

choose two integers $i$ and $j$, then decrease $c_{i,j}$ by $1$. In order to perform this operation, you pay $1$ coin;

choose an integer $i$, then increase $c_{i,j}$ by $1$ for every $j \in [1, n]$ (i. e. increase all elements in the $i$-th row by $1$). When you perform this operation, you receive $1$ coin;

choose an integer $j$, then increase $c_{i,j}$ by $1$ for every $i \in [1, n]$ (i. e. increase all elements in the $j$-th column by $1$). When you perform this operation, you receive $1$ coin.

You have to calculate the minimum number of coins required to transform the matrix $c$ into the zero matrix. Note that all elements of $c$ should be equal to $0$ simultaneously after the operations.


-----Input-----

The first line contains one integer $n$ ($2 \le n \le 2 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^8$).

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($0 \le b_j \le 10^8$).


-----Output-----

Print one integer — the minimum number of coins required to transform the matrix $c$ into the zero matrix.


-----Examples-----

Input
3
1 2 3
2 2 2
Output
2
Input
3
3 1 3
1 1 2
Output
5
Input
2
1 0
2 1
Output
2
Input
2
1 4
2 3
Output
4
Input
4
1 3 3 7
6 9 4 2
Output
29


-----Note-----

In the first example, the matrix looks as follows:

1

1

1

0

0

0

1

1

1

You can turn it into a zero matrix using $2$ coins as follows:

subtract $1$ from the first row, paying $1$ coin;

subtract $1$ from the third row, paying $1$ coin.

In the second example, the matrix looks as follows:

2

2

1

0

0

1

2

2

1

You can turn it into a zero matrix using $5$ coins as follows:

subtract $1$ from the first row, paying $1$ coin;

subtract $1$ from the third row, paying $1$ coin;

subtract $1$ from the third row, paying $1$ coin;

subtract $1$ from $a_{2,3}$, paying $1$ coin;

add $1$ to the third column, receiving $1$ coin;

subtract $1$ from the first row, paying $1$ coin;

subtract $1$ from $a_{2,3}$, paying $1$ coin.

Difficulty: VERY_HARD
Tags: Flows and cuts

Mathematics

Starter code:

Solutions:
from bisect import *
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
a.sort()
b.sort()
sum = [0] * (n + 1)
for i in range(1, n + 1):
	sum[i] = sum[i - 1] + b[i - 1]
ans = 0
for i in range(n):
	j = bisect_right(b, a[i])
	ans += j * a[i] - sum[j]
	ans += sum[n] - sum[j] - (n - j) * a[i]
	ans -= (n - 1) * abs(a[i] - b[i])
print(ans)


from bisect import *
I = lambda : map(int, input().split())
n = int(input())
a = sorted(I())
b = sorted(I())
sum = [0] * (n + 1)
for i in range(1, n + 1):
	sum[i] = sum[i - 1] + b[i - 1]
ans = 0
for i in range(n):
	j = bisect_right(b, a[i])
	ans += j * a[i] - sum[j] + sum[n] - sum[j] - (n - j) * a[i] - (n - 1) * abs(a[i] - b[i])
print(ans)