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. 
from sys import stdin, stdout

MOD = 10**9+7


def fac(n, f):
    f.append(1)

    res = 1
    for i in range(1, n+1):
        res *= i
        res %= MOD
        f.append(res)


def inv(a):
    return bp(a, MOD - 2, MOD)


def bp(x, y, m):
    if y == 0:
        return 1

    p = bp(x, y // 2, m) % m
    p = (p * p) % m

    if y % 2 == 0:
        return p
    else:
        return (x * p) % m


def C(n, c, f):
    p1 = f[n]
    p2 = (f[n-c] * f[c]) % MOD
    invp = inv(p2)
    res = (p1 * invp) % MOD
    #res = p1 // p2

    return res


def cyclic_permutations(n, f):
    res = f[n] - 1
    res %= MOD

    for i in range(1, n):
        res -= C(n-1, i, f)
        res %= MOD

    return res


def cyclic_permutations2(n):
    a = 1
    f = 1
    for i in range(1, n):
        a *= 2
        f *= i
        a %= MOD
        f %= MOD
    f *= n
    f %= MOD
    res = (f - a) % MOD

    return res

n = int(stdin.readline())
#f = []
#fac(n, f)
#ans = cyclic_permutations(n, f)
ans = cyclic_permutations2(n)
stdout.write(str(ans) + '\n')
