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;
long long pw(long long x, long long y) {
  if (y == 0) return 1;
  long long res = pw(x, y / 2);
  res = (res * res) % 1000000007;
  if (y % 2) res = (res * x) % 1000000007;
  return res;
}
long long fact[1000010];
long long dp[1000010];
long long Solve(long long x, int num) {
  if (x == 1) {
    return 1;
  }
  long long &res = dp[x];
  if (res != 0) return res;
  res = 0;
  if (x % 2 == 0) {
    long long z = x / 2;
    long long cnt = 0;
    for (int i = 1; z * i <= n; i += 2) {
      cnt++;
    }
    if (cnt == num) {
      res = fact[cnt];
    } else {
      long long k = Solve(z, num - cnt);
      for (int i = 1; i <= cnt; i++) {
        res = (((((fact[cnt] * pw(fact[cnt - i], 1000000007 - 2)) % 1000000007 *
                  k) %
                 1000000007) *
                ((fact[num - i - 1] * pw(fact[num - 1 - cnt], 1000000007 - 2)) %
                 1000000007)) %
                   1000000007 +
               res) %
              1000000007;
      }
    }
  }
  if (x % 3 == 0) {
    long long z = x / 3;
    long long cnt = 0;
    for (int i = 1; z * i <= n; i++) {
      if (i % 3 == 0) continue;
      cnt++;
    }
    if (cnt == num) {
      res = fact[cnt];
    } else {
      long long k = Solve(z, num - cnt);
      for (int i = 1; i <= cnt; i++) {
        res = (((((fact[cnt] * pw(fact[cnt - i], 1000000007 - 2)) % 1000000007 *
                  k) %
                 1000000007) *
                ((fact[num - i - 1] * pw(fact[num - 1 - cnt], 1000000007 - 2)) %
                 1000000007)) %
                   1000000007 +
               res) %
              1000000007;
      }
    }
  }
  return res;
}
int main() {
  cin >> n;
  fact[0] = 1;
  for (int i = 1; i <= n; i++) {
    fact[i] = (fact[i - 1] * i) % 1000000007;
  }
  long long z = 1;
  while (z <= n) {
    z *= 2;
  }
  z /= 2;
  long long z1 = (z / 2) * 3;
  long long ans = Solve(z, n - 1);
  if (z1 <= n) {
    ans = (ans + Solve(z1, n - 1)) % 1000000007;
  }
  cout << ans << endl;
  return 0;
}
