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;
void debug_out() { cerr << endl; }
template <class T>
ostream& prnt(ostream& out, T v) {
  out << v.size() << '\n';
  for (auto e : v) out << e << ' ';
  return out;
}
template <class T>
ostream& operator<<(ostream& out, vector<T> v) {
  return prnt(out, v);
}
template <class T>
ostream& operator<<(ostream& out, set<T> v) {
  return prnt(out, v);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, map<T1, T2> v) {
  return prnt(out, v);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> p) {
  return out << '(' << p.first << ' ' << p.second << ')';
}
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << H;
  debug_out(T...);
}
const long long N = 1001000;
const long long MOD = 1e9 + 7;
int n, dp[N][22][2];
long long f(long long p2, long long p3) {
  if (p3) return n / ((1 << p2) * 3);
  return n / (1 << p2);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  long long lg = 0, nn = n;
  while (nn) nn /= 2, lg++;
  lg--;
  dp[1][lg][0] = 1;
  if ((1 << (lg - 1)) * 3 <= n) dp[1][lg - 1][1] = 1;
  for (long long i = 1; i < n; i++)
    for (long long x = 0; x <= lg; x++)
      for (long long y = 0; y <= 1; y++) {
        dp[i + 1][x][y] += (1LL * dp[i][x][y] * (f(x, y) - i)) % MOD;
        dp[i][x][y] %= MOD;
        if (x > 0)
          dp[i + 1][x - 1][y] +=
              (1LL * dp[i][x][y] * (f(x - 1, y) - f(x, y))) % MOD,
              dp[i + 1][x - 1][y] %= MOD;
        if (y > 0)
          dp[i + 1][x][y - 1] +=
              (1LL * dp[i][x][y] * (f(x, y - 1) - f(x, y))) % MOD,
              dp[i + 1][x][y - 1] %= MOD;
      }
  cout << dp[n][0][0];
}
