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.
"""
this is a standard python template for codeforces task, repo: github.com/solbiatialessandro/pyComPro/codeforces
"""
stdin = lambda type_ = "int", sep = " ": list(map(eval(type_), raw_input().split(sep)))
joint = lambda sep = " ", *args: sep.join(str(i) if type(i) != list else sep.join(map(str, i)) for i in args)
def iters(): return xrange(int(raw_input()))

def solve(n):
    p = 998244353
    left, right = [-1 for _ in xrange(n + 1)],[-1 for _ in xrange(n + 1)]
    curr = 1
    for i in xrange(n):
        left[i] = curr * (n - i) % p
        curr = left[i]
    curr = 1
    for i in xrange(1, n + 1):
        right[i] = curr * (i) % p
        curr = right[i]
    res = 0
    #print left, right
    for i in xrange(n-1):
        #print i, left[i], right[n-i-1]
        num = left[i] * (right[n - i-1] - 1)
        #print num
        res += num
    return (res + right[-1]) % p

if __name__ == "__main__":
    """the solve(*args) structure is needed for testing purporses"""
    n = int(raw_input())
    print solve(n)