https://codeforces.com/contest/1165/problem/E

题意：重新排序b数组，使得答案最小

题解：排序+贪心

a数组不变

定义

对于排序固定的，中，在ANS中的出现次数与有关，其次数为

其中可以视作定值

定义

所以答案即重新排列Di和Bi的,其中D和B的排序顺序相反

 

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=200000+10;
const int M=100000+10;
const int MOD=998244353;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
ll a[N],b[N];
char str;
struct node{};
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    //scanf("%d",&t);
    //while(t--){
    scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%I64d",&a[i]);
    for(int i=1;i<=n;i++)scanf("%I64d",&b[i]);
    for(int i=1;i<=n;i++)a[i]=(ll)(n-i+1)*i*a[i];
    sort(a+1,a+n+1,greater<ll>() );
    sort(b+1,b+n+1,less<ll>() );
    for(int i=1;i<=n;i++){
        ans=(ans+((a[i]%MOD)*b[i]%MOD)%MOD)%MOD;
    }
    cout<<ans<<endl;
    //}

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}