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>
#pragma GCC optimize("O3")
#pragma GCC target("tune=native")
#pragma GCC optimize("fast-math")
using namespace std;
const int DEBUG = 0;
mt19937 gen(
    (unsigned)chrono::high_resolution_clock::now().time_since_epoch().count());
#pragma comment(linker, "/STACK:16777216")
const int N = (int)1e6 + 20;
const int MOD = (int)1e9 + 7;
int dp[N][20][2];
int n;
int f(int p2, int p3) {
  int x = n / (1 << p2);
  if (p3) x /= 3;
  return x;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.setf(cout.fixed);
  cout.precision(12);
  cin >> n;
  int pow2 = 0;
  while ((1 << pow2) <= n) pow2++;
  --pow2;
  dp[1][pow2][0] = 1;
  if ((1 << (pow2 - 1)) * 3 <= n) dp[1][pow2 - 1][1] = 1;
  for (int i = 1; i < n; i++)
    for (int p2 = 0; p2 <= pow2; p2++)
      for (int p3 = 0; p3 <= 1; p3++) {
        dp[i + 1][p2][p3] =
            (dp[i + 1][p2][p3] + 1LL * dp[i][p2][p3] * (f(p2, p3) - i)) % MOD;
        if (p2 > 0)
          dp[i + 1][p2 - 1][p3] =
              (dp[i + 1][p2 - 1][p3] +
               1LL * dp[i][p2][p3] * (f(p2 - 1, p3) - f(p2, p3))) %
              MOD;
        if (p3 > 0)
          dp[i + 1][p2][p3 - 1] =
              (dp[i + 1][p2][p3 - 1] +
               1LL * dp[i][p2][p3] * (f(p2, p3 - 1) - f(p2, p3))) %
              MOD;
      }
  cout << dp[n][0][0];
  return 0;
}
