A permutation of length n 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).

Consider a permutation p of length n, we build a graph of size n using it as follows: 

  * For every 1 ≤ i ≤ n, find the largest j such that 1 ≤ j < i and p_j > p_i, and add an undirected edge between node i and node j 
  * For every 1 ≤ i ≤ n, find the smallest j such that i < j ≤ n and p_j > p_i, and add an undirected edge between node i and node j 



In cases where no such j exists, we make no edges. Also, note that we make edges between the corresponding indices, not the values at those indices.

For clarity, consider as an example n = 4, and p = [3,1,4,2]; here, the edges of the graph are (1,3),(2,1),(2,3),(4,3).

A permutation p is cyclic if the graph built using p has at least one simple cycle. 

Given n, find the number of cyclic permutations of length n. Since the number may be very large, output it modulo 10^9+7.

Please refer to the Notes section for the formal definition of a simple cycle

Input

The first and only line contains a single integer n (3 ≤ n ≤ 10^6).

Output

Output a single integer 0 ≤ x < 10^9+7, the number of cyclic permutations of length n modulo 10^9+7.

Examples

Input


4


Output


16

Input


583291


Output


135712853

Note

There are 16 cyclic permutations for n = 4. [4,2,1,3] is one such permutation, having a cycle of length four: 4 → 3 → 2 → 1 → 4.

Nodes v_1, v_2, …, v_k form a simple cycle if the following conditions hold: 

  * k ≥ 3. 
  * v_i ≠ v_j for any pair of indices i and j. (1 ≤ i < j ≤ k) 
  * v_i and v_{i+1} share an edge for all i (1 ≤ i < k), and v_1 and v_k share an edge. 
66
 652615
 482331
 336161
 33
 1000000
 79531
 768208
 3
 885131

257415584
 960319213
 722928541
 234634596
 762187807
 23581336
 162141608
 635322133
 2
 329995454

from sys import stdin, stdout
import heapq
import cProfile
from collections import Counter, defaultdict, deque
from functools import reduce
import math


def get_int():
    return int(stdin.readline().strip())


def get_tuple():
    return map(int, stdin.readline().split())


def get_list():
    return list(map(int, stdin.readline().split()))


def power(x, y, p):
    res = 1;  # Initialize result
    x = x % p;  # Update x if it is more
    # than or equal to p
    while (y > 0):

        # If y is odd, multiply
        # x with result
        if (y & 1):
            res = (res * x) % p;

            # y must be even now
        y = y >> 1;  # y = y/2
        x = (x * x) % p;

    return res


# Function to find modular inverse
# of a under modulo p using Fermat's
# method. Assumption: p is prime
def modInverse(a, p):
    return power(a, p - 2, p)


# Returns n! % p using
# Wilson's Theorem
def modFact(n, p):
    # n! % p is 0 if n >= p
    if (p <= n):
        return 0

    # Initialize result as (p-1)!
    # which is -1 or (p-1)
    res = (p - 1)

    # Multiply modulo inverse of
    # all numbers from (n+1) to p
    for i in range(n + 1, p):
        res = (res * modInverse(i, p)) % p
    return res


def is_prime(n):
    """Returns True if n is prime."""
    if n < 4:
        return True
    if n % 2 == 0:
        return False
    if n % 3 == 0:
        return False
    i = 5
    w = 2
    while i * i <= n:
        if n % i == 0:
            return False
        i += w
        w = 6 - w
    return True

m = 10**9 +7
def my_fact(n):
    ans = 1
    for i in range(1,n+1):
        ans = (ans%m * i%m)%m
    return ans

def main():
    n = int(input())
    ans = (my_fact(n) % m - pow(2, n - 1, m) % m + m) % m
    print(ans)


TestCases = False

if TestCases:
    for i in range(get_int()):
        main()
else:
    main()
