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
from __future__ import division, print_function
def main():
    n=int(input())
    l1=list(map(int,input().split()))
    l2=list(map(int,input().split()))
    mod=998244353
    l3=[0]*n
    l2.sort()
    l2=l2[::-1]
    l4=[0]*n
    x=(n*(n+1))//2
    for i in range(0,n):
        y=n-i-1
        l3[i]=x-(i*(i+1))//2-(y*(y+1))//2
    for i in range(0,n):
        l4[i]=l1[i]*l3[i]
    l4.sort()
    s=0
    for i in range(0,n):
        s=(s%mod+l2[i]*l4[i]%mod)%mod
    print(s)

######## Python 2 and 3 footer by Pajenegod and c1729

# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.

# So on cf, use PyPy2 for best string performance.

py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s

    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()

    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s:self.buffer.write(s.encode('ascii'))
            self.read = lambda:self.buffer.read().decode('ascii')
            self.readline = lambda:self.buffer.readline().decode('ascii')


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')

# Cout implemented in Python
import sys
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'

# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
    conv = ord if py2 else lambda x:x
    A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'0' [0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-' [0]: sign = -1
            elif s[i] != b'\r' [0]:
                A.append(sign*numb)
                numb = zero; sign = 1
            i += 1
    except:pass
    if s and s[-1] >= b'0' [0]:
        A.append(sign*numb)
    return A

if __name__== "__main__":
  main()