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;
int n;
ll ans = 0;
ll const P = 1e9 + 7;
ll factorial[(int)1e6 + 2];
void piknie(int m) {
  factorial[0] = 1;
  for (int i = 1; i <= m; i++) factorial[i] = (factorial[i - 1] * (ll)i) % P;
}
ll power(ll a, ll pow) {
  if (pow == 0) return 1;
  ll x = power(a, pow / 2);
  if (pow % 2)
    return (a * ((x * x) % P)) % P;
  else
    return (x * x) % P;
}
ll licz(vector<int>& ciag) {
  ll ret = 1;
  ret = (ret * (n / ciag[0])) % P;
  for (int i = 1; i < ciag.size(); i++) {
    int ro = n / ciag[i] - n / ciag[i - 1];
    ret = (ret * (ll)ro) % P;
  }
  ret = (ret * factorial[n]) % P;
  for (int i = ciag.size() - 1; i >= 1; i--) {
    int ro = n - n / ciag[i - 1];
    ret = (ret * power(ro, P - 2)) % P;
  }
  ret = (ret * power(n, P - 2)) % P;
  return ret;
}
int main() {
  scanf("%d", &n);
  if (n < 3) {
    ans = 1;
  } else {
    int r = (int)(log2(n));
    int potdwa = (1 << r);
    piknie(n);
    vector<int> ciag;
    for (int i = 0; i <= r; i++) {
      ciag.push_back((1 << i));
    }
    reverse(ciag.begin(), ciag.end());
    ans = (ans + licz(ciag)) % P;
    if (3 * potdwa <= 2 * n) {
      for (int i = 1; i <= r; i++) {
        vector<int> krowa;
        for (int j = 0; j < i; j++) {
          krowa.push_back((1 << j));
        }
        for (int j = i; j <= r; j++) {
          krowa.push_back(3 * (1 << (j - 1)));
        }
        reverse(krowa.begin(), krowa.end());
        ans = (ans + licz(krowa)) % P;
      }
    }
  }
  printf("%I64d", ans);
  return 0;
}
