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;
int n;
int dp[2][21][2], maxi, p2[21];
int val(int a, int b) {
  int v = 1;
  if (b == 1) v = 3;
  return (n / (p2[a] * v));
}
int main() {
  int i, j, k;
  ios::sync_with_stdio(false);
  cin >> n;
  p2[0] = 1;
  for (i = 1; i <= 20; i++) p2[i] = (p2[i - 1] * 2) % 1000000007;
  maxi = p2[0];
  i = 0;
  while ((i < 20) && (p2[i + 1] <= n)) i++;
  maxi = i;
  dp[1][maxi][0] = 1;
  if (3LL * p2[(maxi - 1)] <= n) dp[1][maxi - 1][1] = 1;
  for (i = 2; i <= n; i++)
    for (j = 0; j <= maxi; j++) {
      dp[i % 2][j][0] = 0;
      dp[i % 2][j][1] = 0;
      for (k = 0; k <= 1; k++) {
        dp[i % 2][j][k] +=
            ((1LL * val(j, k) - (i - 1)) * dp[1 - i % 2][j][k]) % 1000000007;
        dp[i % 2][j][k] %= 1000000007;
        if (j > 0)
          dp[i % 2][j - 1][k] +=
              ((1LL * val(j - 1, k) - val(j, k)) * (dp[1 - i % 2][j][k])) %
              1000000007;
        dp[i % 2][j - 1][k] %= 1000000007;
        if (k > 0)
          dp[i % 2][j][k - 1] +=
              ((1LL * val(j, k - 1) - val(j, k)) * (dp[1 - i % 2][j][k])) %
              1000000007;
        dp[i % 2][j][k - 1] %= 1000000007;
      }
    }
  cout << dp[n % 2][0][0];
  return 0;
}
