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 double PI = 3.14159265358979323846;
const double eps = (1e-7);
int dcmp(long double x, long double y) {
  if (abs(x - y) < eps) return 0;
  if (x > y) return 1;
  return -1;
}
int n;
const int MOD = 1000000007;
long long add(long long x, long long y) { return (x % MOD + y % MOD) % MOD; }
long long mult(long long x, long long y) { return (x % MOD * y % MOD) % MOD; }
int floorLog(int x) {
  for (int i = 1;; ++i) {
    if ((1 << i) > x) return i - 1;
  }
}
const int MAX_N = 1000005;
int mem[MAX_N][21][2];
int f(int x, int y) {
  int ret = (1 << x);
  if (y) ret *= 3;
  return ret;
}
int countDiv(int x) { return n / x; }
long long solve(int i, int x, int y) {
  if (i == n) return 1;
  int &ret = mem[i][x][y];
  if (~ret) return ret;
  int v = f(x, y);
  ret = 0;
  if (countDiv(v) > i)
    ret = add(ret, mult(solve(i + 1, x, y), (countDiv(v) - i)));
  if (x)
    ret = add(ret, mult(solve(i + 1, x - 1, y), countDiv(v / 2) - countDiv(v)));
  if (y)
    ret = add(ret, mult(solve(i + 1, x, y - 1), countDiv(v / 3) - countDiv(v)));
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  int x = floorLog(n);
  memset(mem, -1, sizeof(mem));
  long long ans = solve(1, x, 0);
  x--;
  if ((1 << x) * 3 <= n) ans = add(ans, solve(1, x, 1));
  cout << ans << '\n';
  return 0;
}
