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;
const int N = 1e6 + 7, mod = 1e9 + 7;
inline int inc(int x, int y) {
  x += y - mod;
  return x + (x >> 31 & mod);
}
inline int del(int x, int y) {
  x -= y;
  return x + (x >> 31 & mod);
}
inline int mul(int x, int y) { return 1ll * x * y % mod; }
inline void Inc(int &x, int y) { x = inc(x, y); }
inline void Del(int &x, int y) { x = del(x, y); }
inline void Mul(int &x, int y) { x = mul(x, y); }
inline void Max(int &x, int y) {
  if (y > x) x = y;
}
inline void Min(int &x, int y) {
  if (y < x) x = y;
}
int n, tot, f[N][22][3], p2[22], p3[3];
inline int read() {
  int f = 1, x = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + (c ^ '0');
    c = getchar();
  }
  return f * x;
}
signed main() {
  n = read();
  tot = log2(n);
  p2[0] = p3[0] = 1;
  for (int i = 1; i <= tot + 1; i++) p2[i] = p2[i - 1] * 2;
  p3[1] = 3, p3[2] = 9;
  f[1][tot][0] = 1;
  if (p2[tot - 1] * 3 <= n) f[1][tot - 1][1] = 1;
  for (int x = 2; x <= n; x++)
    for (int i = 0; i <= tot; i++)
      for (int j = 0; j <= 1; j++) {
        Inc(f[x][i][j], mul(f[x - 1][i][j], n / (p2[i] * p3[j]) - x + 1));
        Inc(f[x][i][j], mul(f[x - 1][i + 1][j],
                            n / (p2[i] * p3[j]) - n / (p2[i + 1] * p3[j])));
        Inc(f[x][i][j], mul(f[x - 1][i][j + 1],
                            n / (p2[i] * p3[j]) - n / (p2[i] * p3[j + 1])));
      }
  printf("%d\n", f[n][0][0]);
  return 0;
}
