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 MAXN = 1000 * 1000;
const long long MOD = 1000 * 1000 * 1000 + 7;
int n, mp = 0, mn = 1;
int a[30], b[30], c[30];
int powers[30] = {1};
long long ans = 1;
void findMaxPower();
long long f(int num, int k);
int main() {
  cin >> n;
  if (n == 1) {
    cout << 1 << endl;
    return 0;
  }
  for (int i = 1; i < 30; i++) powers[i] = powers[i - 1] * 2;
  findMaxPower();
  for (int i = 0; i <= mp; i++) {
    int t = 1;
    while ((2 * t - 1) * powers[i] <= n) {
      a[i]++;
      t++;
    }
  }
  int tn = n;
  for (int i = mp; i >= 0; i--) {
    ans = ans * a[i] % MOD;
    ans = ans * f(tn - 1, a[i] - 1) % MOD;
    tn -= a[i];
  }
  if (!(mn % 2 == 0 && mn / 2 * 3 <= n)) {
    cout << ans << endl;
    return 0;
  }
  for (int i = 0; i < mp; i++) {
    int t = 1;
    while ((2 * t - 1) * powers[i] * 3 <= n) {
      b[i + 1]++;
      t++;
    }
  }
  for (int i = mp; i >= 0; i--) {
    int t = 1;
    while ((2 * t - 1) * powers[i] <= n) {
      if ((2 * t - 1) % 3) c[i]++;
      t++;
    }
    c[i] += c[i + 1];
  }
  for (int j = mp - 1; j >= 0; j--) {
    long long tmp = 1;
    tn = n - 1;
    for (int i = mp - 1; i > j; i--) {
      tmp = tmp * b[i] % MOD;
      tmp = tmp * f(tn - 1, b[i] - 1) % MOD;
      tn -= b[i];
    }
    tmp = tmp * c[j] % MOD;
    tmp = tmp * f(tn - 1, c[j] - 1) % MOD;
    tn -= c[j];
    for (int i = j - 1; i >= 0; i--) {
      tmp = tmp * a[i] % MOD;
      tmp = tmp * f(tn - 1, a[i] - 1) % MOD;
      tn -= a[i];
    }
    ans = (ans + tmp) % MOD;
  }
  cout << ans << endl;
  return 0;
}
void findMaxPower() {
  while (2 * mn <= n) {
    mn *= 2;
    mp++;
  }
}
long long f(int num, int k) {
  long long res = 1;
  for (int i = 0; i < k; i++) res = res * (num - i) % MOD;
  return res;
}
