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;
template <class T>
T gi() {
  T x = 0;
  bool f = 0;
  char c = getchar();
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  if (c == '-') f = 1, c = getchar();
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return f ? -x : x;
}
const int N = 1e6 + 10, Mod = 1e9 + 7;
int f[N][20][2], n, mi[20];
void pls(int &x, int y) {
  x += y;
  if (x >= Mod) x -= Mod;
}
int calc(int x, int y) { return n / (mi[x] * (y ? 3 : 1)); }
int main() {
  n = gi<int>();
  int c = log2(n);
  for (int i = mi[0] = 1; i <= c; i++) mi[i] = 2 * mi[i - 1];
  f[1][c][0] = 1;
  if (pow(2, c - 1) * 3 <= n) f[1][c - 1][1] = 1;
  for (int i = 1; i < n; i++)
    for (int j = 0; j <= c; j++)
      for (int k = 0; k < 2; k++) {
        pls(f[i + 1][j][k], 1ll * f[i][j][k] * (calc(j, k) - i) % Mod);
        if (j)
          pls(f[i + 1][j - 1][k],
              1ll * f[i][j][k] * (calc(j - 1, k) - calc(j, k)) % Mod);
        if (k)
          pls(f[i + 1][j][k - 1],
              1ll * f[i][j][k] * (calc(j, k - 1) - calc(j, k)) % Mod);
      }
  printf("%d\n", f[n][0][0]);
  return 0;
}
