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 sys import stdin
mod=998244353
n=int(stdin.readline().strip())
if n==1:
    print(1)
    exit(0)
"""
global a
a=[i+1 for i in range(n)]
def next_permutation():
    global a

    # Find the largest index i such that a[i] < a[i + 1]. If no such
    # index exists, the permutation is the last permutation
    for i in reversed(range(len(a) - 1)):
        if a[i] < a[i + 1]:
            break  # found
    else:  # no break: not found
        return False  # no next permutation

    # Find the largest index j greater than i such that a[i] < a[j]
    j = next(j for j in reversed(range(i + 1, len(a))) if a[i] < a[j])

    # Swap the value of a[i] with that of a[j]
    a[i], a[j] = a[j], a[i]

    # Reverse sequence from a[i + 1] up to and including the final element a[n]
    a[i + 1:] = reversed(a[i + 1:])
    return True
    
s=[]
s+=tuple(a)
while next_permutation():
    s+=tuple(a)
ans=[]
for i in range(len(s)-n+1):
    sm=0
    aux=[]
    for j in range(i,i+n):
        sm+=s[j]
        aux.append(s[j])
    if sm==sum(a):
        ans.append(tuple(aux))
st=set(ans)
dp=[0 for i in range(n+1)]
for i in st:
    dp[ans.count(i)]+=1
print(dp)
"""

dp=[0 for i in range(n+1)]
dp[1]=n
dp[2]=(n*(n-2))%mod
fact=[1 for i in range(n+1)]
y=n-1
for i in range(1,n+1):
    fact[i]=(fact[i-1]*y)%mod
    y-=1
for i in range(3,n):
    dp[i]=n*(fact[i-2]*(n-i))%mod

ans=0
for i in range(1,n):
    ans=(ans+(i*dp[i]))%mod
print(ans)
