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 long long mod = 1e9 + 7;
int add(long long a, long long b) { return (a + b) % mod; }
int mul(long long a, long long b) { return (a * b) % mod; }
int n;
int dp[1000001][21][2];
int p2[30];
int pow2(int x) {
  if (p2[x] != 0) return p2[x];
  p2[x] = (1 << x);
  return p2[x];
}
int p3[2];
int pow3(int x) {
  if (p3[x] != 0) return p3[x];
  if (x == 0) return 1;
  p3[x] = pow3(x - 1) * 3;
  return p3[x];
}
int main(void) {
  cin >> n;
  int x = log2(n);
  dp[1][x][0] = 1;
  if ((1 << (x - 1)) * 3 <= n) dp[1][x - 1][1] = 1;
  for (int i = 1; i <= n - 1; i++) {
    for (int j = 0; j < 21; j++) {
      for (int k = 0; k < 2; k++) {
        if (dp[i][j][k] == 0) continue;
        int f = n / pow2(j) / pow3(k);
        dp[i + 1][j][k] = add(dp[i + 1][j][k], mul(dp[i][j][k], (f - i)));
        if (j > 0) {
          int ff = n / pow2(j - 1) / pow3(k);
          dp[i + 1][j - 1][k] =
              add(dp[i + 1][j - 1][k], mul(dp[i][j][k], (ff - f)));
        }
        if (k > 0) {
          int ff = n / pow2(j) / pow3(k - 1);
          dp[i + 1][j][k - 1] =
              add(dp[i + 1][j][k - 1], mul(dp[i][j][k], (ff - f)));
        }
      }
    }
  }
  cout << dp[n][0][0] << endl;
}
