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.
// https://codeforces.com/contest/1174/submission/55043857 (rainboy)
import java.io.*;
import java.util.*;

public class CF1174E {
	static final int MD = 1000000007;
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		int[] aa = new int[n + 1];
		for (int i = 1; i <= n; i++)
			aa[i] = i % 2 == 0 ? aa[i / 2] + 1 : 0;
		int k = 0;
		while (1 << k + 1 <= n)
			k++;
		int[] cnt = new int[k + 1];
		int[] cnt3 = new int[k + 1];
		for (int i = 1; i <= n; i++) {
			int a = aa[i];
			cnt[a]++;
			if (i % 3 == 0)
				cnt3[a]++;
		}
		long[] dp = new long[k + 1];
		long[] dp_ = new long[k + 1];
		dp[k] = 1;
		for (int i = 1; i < n; i++) {
			for (int h = 0; h <= k; h++) {
				long x = dp[h];
				if (x != 0) {
					if (h > 0)
						dp_[h - 1] = (dp_[h - 1] + x * cnt[h - 1]) % MD;
					dp_[h] = (dp_[h] + x * (n / (1 << h) - i)) % MD;
				}
			}
			long[] tmp = dp; dp = dp_; dp_ = tmp;
			Arrays.fill(dp_, 0);
		}
		long ans = dp[0];
		if (3 << k - 1 <= n) {
			Arrays.fill(dp, 0);
			long[] dq = new long[k];
			long[] dq_ = new long[k];
			dq[k - 1] = 1;
			for (int i = 1; i < n; i++) {
				for (int h = 0; h < k; h++) {
					long x = dq[h];
					if (x != 0) {
						if (h > 0)
							dq_[h - 1] = (dq_[h - 1] + x * cnt3[h - 1]) % MD;
						dq_[h] = (dq_[h] + x * (n / (3 << h) - i)) % MD;
						dp_[h] = (dp_[h] + x * (n / (1 << h) - n / (3 << h))) % MD;
					}
					x = dp[h];
					if (x != 0) {
						if (h > 0)
							dp_[h - 1] = (dp_[h - 1] + x * cnt[h - 1]) % MD;
						dp_[h] = (dp_[h] + x * (n / (1 << h) - i)) % MD;
					}
				}
				long[] tmp = dp; dp = dp_; dp_ = tmp; tmp = dq; dq = dq_; dq_ = tmp;
				Arrays.fill(dp_, 0);
				Arrays.fill(dq_, 0);
			}
			ans = (ans + dp[0]) % MD;
		}
		System.out.println(ans);
	}
}
