Let n be an integer. Consider all permutations on integers 1 to n in lexicographic order, and concatenate them into one big sequence p. For example, if n = 3, then p = [1, 2, 3, 1, 3, 2, 2, 1, 3, 2, 3, 1, 3, 1, 2, 3, 2, 1]. The length of this sequence will be n ⋅ n!.

Let 1 ≤ i ≤ j ≤ n ⋅ n! be a pair of indices. We call the sequence (p_i, p_{i+1}, ..., p_{j-1}, p_j) a subarray of p. Its length is defined as the number of its elements, i.e., j - i + 1. Its sum is the sum of all its elements, i.e., ∑_{k=i}^j p_k. 

You are given n. Find the number of subarrays of p of length n having sum (n(n+1))/(2). Since this number may be large, output it modulo 998244353 (a prime number). 

Input

The only line contains one integer n (1 ≤ n ≤ 10^6), as described in the problem statement.

Output

Output a single integer — the number of subarrays of length n having sum (n(n+1))/(2), modulo 998244353.

Examples

Input


3


Output


9


Input


4


Output


56


Input


10


Output


30052700

Note

In the first sample, there are 16 subarrays of length 3. In order of appearance, they are:

[1, 2, 3], [2, 3, 1], [3, 1, 3], [1, 3, 2], [3, 2, 2], [2, 2, 1], [2, 1, 3], [1, 3, 2], [3, 2, 3], [2, 3, 1], [3, 1, 3], [1, 3, 1], [3, 1, 2], [1, 2, 3], [2, 3, 2], [3, 2, 1]. 

Their sums are 6, 6, 7, 6, 7, 5, 6, 6, 8, 6, 7, 5, 6, 6, 7, 6. As (n(n+1))/(2) = 6, the answer is 9.
#from itertools import permutations 
#from sympy import factorial
# l = list(permutations(range(1, n+1))) 
# a=[]
# print len(l)
# for i in l:
# 	a=a+list(i)
# pre=[0,a[0]]
# #print a
# for i in range(1,len(a)):
# 	pre.append(a[i]+pre[i])
# count=0	
# for i in range(n,len(pre)):
# 	f=(n*(n+1))/2
# 	m=pre[i]-pre[i-n]
# 	#print pre[i],pre[i-n]
# 	if(m==f):
# 		count=count+1
# #print pre
# print count
# print factorial(n)*(n-2)-count
n=int(raw_input())
mod=998244353
fact=[1,1]
for i in range(2,10**6+3):
	fact.append((fact[i-1]*i)%mod)
a=[3,8]
for i in range(1,10**6+10):
	a.append(((i+4)*(a[i]-1)%mod)%mod)
if(n<3):
	print n
else:
	print ((fact[n]*(n-2))%mod+a[n-3])%mod		
