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 maximize(int &a, int b) { a = max(a, b); }
const long long int modulo = 1e9 + 7;
void add(int &a, long long int b) { a = (a + b) % modulo; }
int n;
int memcompute[20][2][1000005];
int expo3[2];
int compute(int x, int y, long long int available) {
  int &mem = memcompute[x][y][available];
  if (mem != -1) return mem;
  mem = 0;
  if (not x and not y and not available) {
    mem = 1;
    return mem;
  }
  if (x) {
    long long int extra =
        n / ((1 << (x - 1)) * expo3[y]) - n / ((1 << x) * expo3[y]);
    add(mem, extra * compute(x - 1, y, available + extra - 1));
  }
  if (y) {
    long long int extra =
        n / ((1 << x) * expo3[y - 1]) - n / ((1 << x) * expo3[y]);
    add(mem, extra * compute(x, y - 1, available + extra - 1));
  }
  if (available) add(mem, compute(x, y, available - 1) * available);
  return mem;
}
int sol;
int main() {
  cin >> n;
  for (int i = 0; i < 20; i++)
    for (int j = 0; j < 2; j++)
      for (int k = 0; k <= n; k++) memcompute[i][j][k] = -1;
  expo3[0] = 1;
  expo3[1] = 3;
  int e = 0;
  while ((1 << (e + 1)) <= n) e++;
  add(sol, compute(e, 0, 0));
  if (e and (1 << (e - 1)) * 3 <= n) add(sol, compute(e - 1, 1, 0));
  cout << sol << endl;
}
