Recall that the permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).

A sequence a is a subsegment of a sequence b if a can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end. We will denote the subsegments as [l, r], where l, r are two integers with 1 ≤ l ≤ r ≤ n. This indicates the subsegment where l-1 elements from the beginning and n-r elements from the end are deleted from the sequence.

For a permutation p_1, p_2, …, p_n, we define a framed segment as a subsegment [l,r] where max\\{p_l, p_{l+1}, ..., p_r\} - min\\{p_l, p_{l+1}, ..., p_r\} = r - l. For example, for the permutation (6, 7, 1, 8, 5, 3, 2, 4) some of its framed segments are: [1, 2], [5, 8], [6, 7], [3, 3], [8, 8]. In particular, a subsegment [i,i] is always a framed segments for any i between 1 and n, inclusive.

We define the happiness of a permutation p as the number of pairs (l, r) such that 1 ≤ l ≤ r ≤ n, and [l, r] is a framed segment. For example, the permutation [3, 1, 2] has happiness 5: all segments except [1, 2] are framed segments.

Given integers n and m, Jongwon wants to compute the sum of happiness for all permutations of length n, modulo the prime number m. Note that there exist n! (factorial of n) different permutations of length n.

Input

The only line contains two integers n and m (1 ≤ n ≤ 250 000, 10^8 ≤ m ≤ 10^9, m is prime).

Output

Print r (0 ≤ r < m), the sum of happiness for all permutations of length n, modulo a prime number m.

Examples

Input


1 993244853


Output


1


Input


2 993244853


Output


6


Input


3 993244853


Output


32


Input


2019 993244853


Output


923958830


Input


2020 437122297


Output


265955509

Note

For sample input n=3, let's consider all permutations of length 3:

  * [1, 2, 3], all subsegments are framed segment. Happiness is 6. 
  * [1, 3, 2], all subsegments except [1, 2] are framed segment. Happiness is 5. 
  * [2, 1, 3], all subsegments except [2, 3] are framed segment. Happiness is 5. 
  * [2, 3, 1], all subsegments except [2, 3] are framed segment. Happiness is 5. 
  * [3, 1, 2], all subsegments except [1, 2] are framed segment. Happiness is 5. 
  * [3, 2, 1], all subsegments are framed segment. Happiness is 6. 



Thus, the sum of happiness is 6+5+5+5+5+6 = 32.
from __future__ import division, print_function
def main():
    n, m = map(int,input().split())
    ans = 0
    fact = [1]
    for i in range(1,n+1):
        fact.append((fact[-1]*i)%m)
    
    for i in range(1,n+1):
        x = n - i + 1
        ans = (ans+fact[i]*x * fact[n-i+1])%m
    print(ans)
    
######## 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()