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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
long long N = 1e9 + 7, M = 1000000;
long long n, m, t, k, k2, f, mi = 2e9, x, y, z, ma, x3, x2, y2, l, r, i, j, ans,
                             a[1110000][3], b[410000], d[1110000], dp[410000],
                             p[410000];
string s, s2;
vector<long long> v, v2, v3;
long long voz(long long x, long long y) {
  if (y == 1) return x;
  if (y == 0) return 1ll;
  long long t = (voz(x, y / 2));
  if (y % 2) return (t * t % N * x) % N;
  return (t * t) % N;
}
long long F(long long x, long long y) {
  return d[x + y] * voz(d[y], N - 2) % N;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  m = n;
  while (m > 0) {
    k++;
    m /= 2;
  }
  for (i = 1; i <= n; i++) {
    f = 0;
    for (j = 1; j <= 20; j++) {
      if ((i % (1 << j)) == 0)
        f = j;
      else
        break;
    }
    x = (i % 3 == 0);
    a[f][x]++;
  }
  d[0] = 1;
  for (i = 1; i <= n; i++) d[i] = (d[i - 1] * i) % N;
  k--;
  for (j = 0; j <= k; j++) {
    t = 1;
    x = 0;
    for (i = 0; i < j; i++) {
      t = (t * (a[i][0] + a[i][1]) % N * F(a[i][0] + a[i][1] - 1, x)) % N;
      x += a[i][0] + a[i][1];
    }
    l = 0;
    for (i = j; i <= k; i++) l += a[i][0];
    t = (t * (l) % N * F(l - 1, x)) % N;
    x += l;
    for (i = j; i < k; i++) {
      t = (t * (a[i][1]) % N * F(a[i][1] - 1, x)) % N;
      x += a[i][1];
    }
    ans = (ans + t) % N;
  }
  cout << ans;
}
