Let's define a function f(p) on a permutation p as follows. Let g_i be the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of elements p_1, p_2, ..., p_i (in other words, it is the GCD of the prefix of length i). Then f(p) is the number of distinct elements among g_1, g_2, ..., g_n.

Let f_{max}(n) be the maximum value of f(p) among all permutations p of integers 1, 2, ..., n.

Given an integers n, count the number of permutations p of integers 1, 2, ..., n, such that f(p) is equal to f_{max}(n). Since the answer may be large, print the remainder of its division by 1000 000 007 = 10^9 + 7.

Input

The only line contains the integer n (2 ≤ n ≤ 10^6) — the length of the permutations.

Output

The only line should contain your answer modulo 10^9+7.

Examples

Input


2


Output


1

Input


3


Output


4

Input


6


Output


120

Note

Consider the second example: these are the permutations of length 3:

  * [1,2,3], f(p)=1. 
  * [1,3,2], f(p)=1. 
  * [2,1,3], f(p)=2. 
  * [2,3,1], f(p)=2. 
  * [3,1,2], f(p)=2. 
  * [3,2,1], f(p)=2. 



The maximum value f_{max}(3) = 2, and there are 4 permutations p such that f(p)=2.
import sys
range = xrange
input = raw_input

MOD = 10**9 + 7
def fac(x):
    i = 1
    for j in range(2 , x + 1):
        i = i * j % MOD
    return i

fac = [1]
for i in range(1, 10**6 + 10):
    fac.append(int(fac[-1] * i % MOD))

n = int(input())
def ways(prevg, g):
    cur = n//g - n//prevg if prevg else n//g
    left = n - n//g

    a = cur
    for i in range(left + 1, left + cur):
        a = a * i % MOD
    #a = fac[cur] * pow(left+1, cur - 1, MOD)
    b = +(g == 1)
    for x in 2,3:
        if g % x == 0:
            b += ways(g, g // x)
    return a * b

a = 1
while a * 2 <= n:
    a *= 2

ans = ways(0, a)
if a //2 * 3 <= n:
    ans += ways(0, a //2 * 3)

print ans % MOD
