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;
signed long long mo = 1000000007;
signed long long N;
const int NUM_ = 1400001;
static signed long long fact[NUM_ + 1], factr[NUM_ + 1], inv[NUM_ + 1];
signed long long comb(signed long long N_, signed long long C_) {
  if (fact[0] == 0) {
    inv[1] = fact[0] = factr[0] = 1;
    for (int i = 2; i <= NUM_; ++i) inv[i] = inv[mo % i] * (mo - mo / i) % mo;
    for (int i = 1; i <= NUM_; ++i)
      fact[i] = fact[i - 1] * i % mo, factr[i] = factr[i - 1] * inv[i] % mo;
  }
  if (C_ < 0 || C_ > N_) return 0;
  return factr[C_] * fact[N_] % mo * factr[N_ - C_] % mo;
}
signed long long hoge(int cur, vector<int> V) {
  int sum = 0;
  signed long long ret = 1;
  for (auto& v : V) {
    int n = N / cur - sum;
    ret = ret * comb(N - sum - 1, n - 1) % mo * fact[n] % mo;
    sum += n;
    cur /= v;
  }
  ret = ret * fact[N - sum] % mo;
  return ret;
}
void solve() {
  int i, j, k, l, r, x, y;
  string s;
  comb(1, 1);
  cin >> N;
  x = y = 0;
  while (1 << (x + 1) <= N) x++;
  while (3 << (y + 1) <= N) y++;
  signed long long ret = 0;
  vector<int> V;
  signed long long sum = 0;
  for (i = 0; i < (x); i++) V.push_back(2);
  ret += hoge(1 << x, V);
  if (y + 1 == x) {
    V.clear();
    for (i = 0; i < (y); i++) V.push_back(2);
    V.push_back(3);
    do {
      ret += hoge(3 << y, V);
    } while (next_permutation((V.begin()), (V.end())));
  }
  cout << ret % mo << endl;
}
int main(int argc, char** argv) {
  string s;
  int i;
  if (argc == 1) ios::sync_with_stdio(false), cin.tie(0);
  for (i = 0; i < (argc - 1); i++) s += argv[i + 1], s += '\n';
  for (i = 0; i < (s.size()); i++) ungetc(s[s.size() - 1 - i], stdin);
  cout.tie(0);
  solve();
  return 0;
}
