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A12592. Ehab and the Expected GCD Problem

编程题 普及/提高-

题目描述

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 $1000\,000\,007 = 10^9 + 7$ .

输入格式

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

输出格式

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

输入输出样例

输入 #1
2
输出 #1
1
输入 #2
3
输出 #2
4
输入 #3
6
输出 #3
120

说明/提示

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 $f(p)=2$ .
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