Two Arrays and Sum of Functions, Contribution Technique

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define all(x) (x).begin(), (x).end()
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pair<int, int>> vpi;
typedef pair<int, int> pi;
#define f first
#define s second
#define pb push_back
#define endl "\n"
#define yes cout << "YES" << endl
#define no cout << "NO" << endl
#define init(x, a) memset(x, a, sizeof(x))
const int mod1 = 1e9 + 7, mod2 = 998244353, INF = 2e18, N = 2e5 + 5, L = 19;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
// -----------------------------------------------------------------------------

void solve()
{
    int n;
    cin >> n;
    vi b(n);
    vpi a(n);
    for (int i = 0; i < n; i++)
    {
        int val;
        cin >> val;
        // important to make it mod
        a[i] = {val * 1LL * (n - i) * (i + 1), i}; // contribution
    }
    for (int i = 0; i < n; i++)
    {
        cin >> b[i];
    }
    sort(all(a), greater<>());
    sort(all(b));
    int sum = 0;
    for (int i = 0; i < n; i++)
    {
        sum = (sum + ((a[i].f % mod2) * b[i]) % mod2) % mod2;
    }
    cout << sum << endl;
    return;
}

signed main()
{
    // __START__;
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t = 1;
    // cin >> t;
    while (t--)
    {
        solve();
    }
    // __END__;
    return 0;
}