human
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. 
Using python3 can you solve the prior task?
 Write me a Python code to 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?
assistant
#Here is Python code to 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 
.

from sys import stdin

def inp():
    return stdin.buffer.readline().rstrip().decode('utf8')

def itg():
    return int(stdin.buffer.readline())

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

# ############################## import
from bisect import bisect_left, bisect_right

MOD = int(1e9) + 7
LOAD = int(1e6) + 3

class Sequential:
    def __init__(self, data=None):
        if data is None:
            data = [1]
        self.data = data

    def __getitem__(self, item):
        if isinstance(item, slice):
            last = item.stop - 1
        else:
            last = item
        if last > LOAD:
            return self.__call__(item)
        if len(self) <= last:
            self.resize(last + 1)
        return self.data[item]

    def resize(self, n):
        """ resize data to n """
        del self.data[n:]
        length = len(self)
        self.data.extend([None] * (n - length))
        for i in range(length, n):  # modify here if you need the index
            self.data[i] = self.next(self.data[i - 1])

    def resize_with_value(self, value):
        """ extend data until data[-1] >= value"""
        if self.data[-1] < value:
            while True:
                self.data.append(self.next(self.data[-1]))
                if self.data[-1] >= value:
                    return self.data[-1]

    @staticmethod
    def next(a_n):
        return NotImplemented

    def prev(self, a_n):
        return self.lower(a_n)

    def bisect_left(self, value):
        """ return smallest i s.t. a(i) >= value"""
        self.resize_with_value(value)
        return bisect_left(self.data, value)

    def bisect_right(self, value):
        """ return smallest i s.t. a(i) > value"""
        self.resize_with_value(value)
        return bisect_right(self.data, value)

    def ceil(self, value):
        return self.data[self.bisect_left(value)]

    def higher(self, value):
        index = self.bisect_left(value)
        ceil = self.data[index]
        return self[index + 1] if ceil == value...