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 <typename T>
void read(T &t) {
  t = 0;
  char ch = getchar();
  int f = 1;
  while ('0' > ch || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  do {
    (t *= 10) += ch - '0';
    ch = getchar();
  } while ('0' <= ch && ch <= '9');
  t *= f;
}
const int mod = (1e9) + 7;
const int maxn = (1e6) + 5;
int n, dp[maxn][21][3];
int m2[21], m3[3];
int f(int x, int y) { return n / (m2[x] * m3[y]); }
void add(int &d, int x, int y) {
  d += (long long)x * y % mod;
  if (d >= mod) d -= mod;
}
int main() {
  read(n);
  int cnt = 0;
  while ((1 << cnt) <= n) cnt++;
  cnt--;
  for (int i = 0; i <= cnt; i++) m2[i] = 1 << i;
  m3[0] = 1;
  m3[1] = 3;
  dp[1][cnt][0] = 1;
  if (f(cnt - 1, 1)) dp[1][cnt - 1][1] = 1;
  for (int i = 1; i <= n; i++) {
    for (int x = 0; x <= cnt; x++)
      for (int y = 0; y <= 1; y++) {
        add(dp[i + 1][x][y], dp[i][x][y], f(x, y) - i);
        if (x) add(dp[i + 1][x - 1][y], dp[i][x][y], f(x - 1, y) - f(x, y));
        if (y) add(dp[i + 1][x][y - 1], dp[i][x][y], f(x, y - 1) - f(x, y));
      }
  }
  printf("%d\n", dp[n][0][0]);
  return 0;
}
