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 mod = 1e9 + 7;
const int N = 1e6 + 5;
int n, fact[N], memo[22][2][N];
int f(int x, int y) {
  int dem = (1 << x);
  if (y) dem *= 3;
  return n / dem;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  fact[0] = 1;
  for (int i = 1; i < N; ++i) {
    fact[i] = (fact[i - 1] * 1LL * i) % mod;
  }
  cin >> n;
  if (n <= 2) {
    cout << 1;
    return 0;
  }
  int mxPow2 = 0;
  while ((1 << mxPow2) <= n) ++mxPow2;
  int pow2 = 1 << (mxPow2 - 1);
  assert(pow2 * 2 > n);
  memo[mxPow2 - 1][0][1] = 1;
  if (pow2 / 2 * 3 <= n) {
    assert(mxPow2 - 2 >= 0);
    memo[mxPow2 - 2][1][1] = 1;
  }
  for (int i = 1; i <= n; ++i) {
    for (int x = 0; x < mxPow2; ++x) {
      for (int y = 0; y <= 1; ++y) {
        int &curr = memo[x][y][i];
        if (f(x, y) >= i - 1) {
          curr += memo[x][y][i - 1] * 1LL * (f(x, y) - (i - 1)) % mod;
          if (curr >= mod) curr -= mod;
        }
        if (x + 1 < mxPow2) {
          curr += (memo[x + 1][y][i - 1] * 1LL * (f(x, y) - f(x + 1, y))) % mod;
          if (curr >= mod) curr -= mod;
        }
        if (y == 0) {
          curr += (memo[x][y + 1][i - 1] * 1LL * (f(x, y) - f(x, y + 1))) % mod;
          if (curr >= mod) curr -= mod;
        }
        assert(0 <= curr && curr < mod);
      }
    }
  }
  cout << memo[0][0][n];
}
