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;
const int MX = 1000005, mod = 1000000007;
int n, frac[MX], inv[MX], d[23], num[23];
int work1(int vp2) {
  for (int i = 0; i <= vp2; i++) d[i] = 1 << i;
  for (int i = 0; i <= vp2; i++) num[i] = n / d[i];
  for (int i = 0; i < vp2; i++) num[i] -= num[i + 1];
  LL ans = frac[n];
  for (int i = 0; i <= vp2; i++) ans = ans * num[i] % mod;
  for (int i = 0, l = 0; i <= vp2; i++) l += num[i], ans = ans * inv[l] % mod;
  return ans;
}
int work2(int k, int vp2) {
  if (k > n || vp2 < 0) return 0;
  LL ans = 0;
  for (int cas = 0; cas <= vp2; cas++) {
    for (int i = 0; i <= cas; i++) d[i] = 1 << i;
    d[cas + 1] = d[cas] * 3;
    for (int i = cas + 1; i <= vp2; i++) d[i + 1] = d[i] << 1;
    for (int i = 0; i <= vp2 + 1; i++) num[i] = n / d[i];
    for (int i = 0; i <= vp2; i++) num[i] -= num[i + 1];
    LL tmp = frac[n];
    for (int i = 0; i <= vp2 + 1; i++) tmp = tmp * num[i] % mod;
    for (int i = 0, l = 0; i <= vp2 + 1; i++)
      l += num[i], tmp = tmp * inv[l] % mod;
    ans += tmp;
  }
  return ans % mod;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  frac[0] = 1;
  for (LL i = 1; i <= n; i++) frac[i] = frac[i - 1] * i % mod;
  inv[1] = 1;
  for (int i = 2; i <= n; i++) inv[i] = LL(mod - mod / i) * inv[mod % i] % mod;
  int pw2 = 1, d = 0;
  while (pw2 <= n) pw2 <<= 1, ++d;
  pw2 >>= 1, --d;
  cout << (work1(d) + work2(pw2 / 2 * 3, d - 1)) % mod << "\n";
  return 0;
}
