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;
using ll = long long;
const ll mod = 1000000007;
ll dp[2][21][2];
int n;
int twos, cnt;
ll calc(int x, int y) { return n / ((y ? 3 : 1) << x); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  memset(dp, 0, sizeof(dp));
  cin >> n;
  for (twos = 1; twos <= n; twos <<= 1, cnt++)
    ;
  cnt -= 1;
  dp[0][cnt][0] = 1;
  dp[0][cnt - 1][1] = ((3 << cnt) <= 2 * n);
  for (int pos = 1; pos < n; pos++) {
    int k = pos & 1;
    for (int x = 0; x < cnt; x++) {
      for (int y = 0; y < 2; y++) {
        ll& ret = dp[k][x][y];
        if (y == 0) {
          ret += dp[1 - k][x][y + 1] * (calc(x, y) - calc(x, y + 1));
          ret %= mod;
        }
        if (x + 1 <= cnt) {
          ret += dp[1 - k][x + 1][y] * (calc(x, y) - calc(x + 1, y));
          ret %= mod;
        }
        if (calc(x, y) > pos) {
          ret += dp[1 - k][x][y] * (calc(x, y) - pos);
          ret %= mod;
        }
      }
    }
    memset(dp[1 - k], 0, sizeof(dp[1 - k]));
  }
  cout << dp[(n - 1) & 1][0][0];
}
