You are given two arrays a and b, both of length n.

Let's define a function f(l, r) = ∑_{l ≤ i ≤ r} a_i ⋅ b_i.

Your task is to reorder the elements (choose an arbitrary order of elements) of the array b to minimize the value of ∑_{1 ≤ l ≤ r ≤ n} f(l, r). Since the answer can be very large, you have to print it modulo 998244353. Note that you should minimize the answer but not its remainder.

Input

The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a and b.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6), where a_i is the i-th element of a.

The third line of the input contains n integers b_1, b_2, ..., b_n (1 ≤ b_j ≤ 10^6), where b_j is the j-th element of b.

Output

Print one integer — the minimum possible value of ∑_{1 ≤ l ≤ r ≤ n} f(l, r) after rearranging elements of b, taken modulo 998244353. Note that you should minimize the answer but not its remainder.

Examples

Input


5
1 8 7 2 4
9 7 2 9 3


Output


646


Input


1
1000000
1000000


Output


757402647


Input


2
1 3
4 2


Output


20
#include <bits/stdc++.h>
using namespace std;
int gcdExtended(int a, int b, int* x, int* y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  int x1, y1;
  int gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
long long int modInverse(int a, int m) {
  int x, y;
  int g = gcdExtended(a, m, &x, &y);
  long long int res = ((long long int)x % m + (long long int)m) % m;
  return res;
}
long long int poww(long long int x, long long int n) {
  if (n == 0)
    return 1;
  else if (n == 1)
    return x;
  else if (n % 2 == 0)
    return poww(x * x, n / 2);
  else
    return x * poww(x * x, (n - 1) / 2);
}
long long int modularExponentiation(long long int x, long long int n,
                                    long long int M) {
  long long int result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % M;
    x = (x * x) % M;
    n = n / 2;
  }
  return result;
}
template <typename T>
T maxx(T a, T b) {
  if (a < b) return b;
  return a;
}
template <typename T, typename... Args>
T maxx(T a, T b, Args... args) {
  return (maxx(maxx(a, b), args...));
}
template <typename T>
T minn(T a, T b) {
  if (a < b) return a;
  return b;
}
template <typename T, typename... Args>
T minn(T a, T b, Args... args) {
  return (minn(minn(a, b), args...));
}
template <typename T>
void read(T& a) {
  cin >> a;
}
template <typename T, typename... Args>
void read(T& a, Args&... args) {
  cin >> a;
  read(args...);
}
template <typename T>
void write(T a) {
  cout << a << ' ';
}
template <typename T, typename... Args>
void write(T a, Args... args) {
  cout << a << ' ';
  write(args...);
}
inline long long int min(long long int a, long long int b) {
  if (a < b)
    return a;
  else
    return b;
}
inline long long int max(long long int a, long long int b) {
  if (a < b)
    return b;
  else
    return a;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  if (b == 0) return a;
  if (a == b) return a;
  if (a > b)
    return gcd(a - b, b);
  else
    return gcd(a, b - a);
}
long long int ncr(int n, int r) {
  long long int ans = 1;
  r = min(r, n - r);
  for (int i = 0; i < r; i++) {
    ans = ((ans * ((n - i) % 1000000007)) % 1000000007 *
           modInverse(i + 1, 1000000007)) %
          1000000007;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    long long int n, m, k;
    cin >> n;
    vector<long long int> a(n);
    vector<int> b(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    ;
    for (int i = 0; i < n; i++) cin >> b[i];
    ;
    vector<long long int> c(n);
    long long int M = 998244353;
    for (int i = 1; i < n + 1; i++) {
      c[i - 1] = (((long long int)i * (n - i + 1)) % M * a[i - 1] % M) % M;
    }
    sort((c).begin(), (c).end());
    sort((b).begin(), (b).end(), greater<int>());
    k = 0;
    for (int i = 0; i < n; i++) {
      k += (c[i] * b[i]) % M;
      k %= M;
    }
    cout << k;
  }
  return 0;
}
