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;
template <class T>
bool maximize(T &x, const T &y) {
  if (x < y)
    x = y;
  else
    return false;
  return true;
}
const int mod = 1e9 + 7;
const int N = 1e6 + 7;
int n, Fmax, m;
int f[2][25][2], cnt[25][2];
void add(int &x, long long y) {
  y %= mod;
  (x += y) %= mod;
}
void Init() {
  int cur2 = 1, cur3 = 1;
  for (int i = 0; i <= m; i++) {
    cur3 = 1;
    for (int j = 0; j <= 1; j++) {
      if (n / cur2 >= cur3) {
        cnt[i][j] = n / (cur2 * cur3);
        maximize(Fmax, i + j);
      }
      cur3 *= 3;
    }
    cur2 *= 2;
  }
}
void DP() {
  int t0 = 0, t1 = 1;
  for (int i = 0; i <= 1; i++) f[0][Fmax - i][i] = cnt[Fmax - i][i];
  for (int i = 2; i <= n; i++) {
    memset(f[t1], 0, sizeof(f[t1]));
    for (int j = 0; j <= Fmax; j++)
      for (int y = 0; y <= 1; y++) {
        int x = j - y;
        if (x < 0) continue;
        if (cnt[x][y] >= i)
          add(f[t1][x][y], 1ll * f[t0][x][y] * (cnt[x][y] - i + 1));
        if (y > 0)
          add(f[t1][x][y - 1], 1ll * f[t0][x][y] * (cnt[x][y - 1] - cnt[x][y]));
        if (x > 0)
          add(f[t1][x - 1][y], 1ll * f[t0][x][y] * (cnt[x - 1][y] - cnt[x][y]));
      }
    t0 ^= 1;
    t1 ^= 1;
  }
  cout << f[t0][0][0];
}
int main() {
  if (fopen("exam.inp", "r"))
    freopen("exam.inp", "r", stdin), freopen("exam.out", "w", stdout);
  cin >> n;
  m = log2(n);
  Init();
  DP();
}
