#P1174E. Ehab and the Expected GCD Problem

    ID: 2466 Type: RemoteJudge 2000ms 256MiB Tried: 0 Accepted: 0 Difficulty: (None) Uploaded By: Tags>combinatoricsdpmathnumber theory*2500

Ehab and the Expected GCD Problem

No submission language available for this problem.

Description

Let's define a function f(p)f(p) on a permutation pp as follows. Let gig_i be the greatest common divisor (GCD) of elements p1p_1, p2p_2, ..., pip_i (in other words, it is the GCD of the prefix of length ii). Then f(p)f(p) is the number of distinct elements among g1g_1, g2g_2, ..., gng_n.

Let fmax(n)f_{max}(n) be the maximum value of f(p)f(p) among all permutations pp of integers 11, 22, ..., nn.

Given an integers nn, count the number of permutations pp of integers 11, 22, ..., nn, such that f(p)f(p) is equal to fmax(n)f_{max}(n). Since the answer may be large, print the remainder of its division by 1000000007=109+71000\,000\,007 = 10^9 + 7.

The only line contains the integer nn (2n1062 \le n \le 10^6) — the length of the permutations.

The only line should contain your answer modulo 109+710^9+7.

Input

The only line contains the integer nn (2n1062 \le n \le 10^6) — the length of the permutations.

Output

The only line should contain your answer modulo 109+710^9+7.

Samples

Sample Input 1

2

Sample Output 1

1

Sample Input 2

3

Sample Output 2

4

Sample Input 3

6

Sample Output 3

120

Note

Consider the second example: these are the permutations of length 33:

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

The maximum value fmax(3)=2f_{max}(3) = 2, and there are 44 permutations pp such that f(p)=2f(p)=2.