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;
int main() {
  const int64_t mod = 1000000007;
  auto mul = [&](int64_t a, int64_t b) { return a * b % mod; };
  int n, k = 0;
  cin >> n;
  for (int m = n >> 1; m; m >>= 1) ++k;
  map<int, int64_t> dp0;
  dp0[(1 << k) * 3] = 1;
  for (int i = 0; i != n; ++i) {
    map<int, int64_t> dp1;
    for (const auto& dp : dp0) {
      int cnt = n / dp.first;
      if (i < cnt) dp1[dp.first] += mul(dp.second, cnt - i);
      if (dp.first % 2 == 0)
        dp1[dp.first / 2] += mul(dp.second, n / (dp.first / 2) - cnt);
      if (dp.first % 3 == 0)
        dp1[dp.first / 3] += mul(dp.second, n / (dp.first / 3) - cnt);
    }
    for (auto& dp : dp1) dp.second %= mod;
    dp0 = move(dp1);
  }
  cout << dp0[1] << endl;
}
