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.
#include <bits/stdc++.h>
using namespace std;
long long modilo = 1000000007;
long long n;
long long ff(vector<long long> e) {
  vector<long long> c;
  long long ss = 1, b = 0;
  for (int i = 0; i < e.size(); i++) {
    c.push_back((n / e[i]) - b);
    ss *= ((n / e[i]) - b);
    ss %= modilo;
    b = (n / e[i]);
  }
  long long xz = 1;
  for (int i = (int)e.size() - 1; i >= 0; i--) {
    for (int j = 0; j < c[i] - 1; j++) {
      ss *= xz;
      ss %= modilo;
      xz++;
    }
    xz++;
  }
  return ss;
}
int main() {
  cin >> n;
  vector<long long> v;
  v.push_back(1);
  while (v[v.size() - 1] * 2LL <= n) v.push_back(v[v.size() - 1] * 2LL);
  reverse(v.begin(), v.end());
  long long p = ff(v);
  long long ss = p;
  if (3LL * v[0] <= 2LL * n) {
    for (int i = 1; i <= (int)v.size() - 1; i++) {
      vector<long long> v2 = v;
      for (int j = 0; j < i; j++) v2[j] = 3LL * v2[j] / 2LL;
      long long p2 = ff(v2);
      ss += p2;
      ss %= modilo;
    }
  }
  cout << ss << endl;
}
