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, m = 1e9 + 7, dp[1000002][20][2], p1[20], p2[5];
int solve1(int x, int y) { return n / (p1[x] * p2[y]); }
int solve(int i, int x, int y) {
  if (i + x + y == 0) return 1;
  int &ans = dp[i][x][y];
  if (ans != -1) return ans;
  if (x + y == 1) {
    if (y)
      ans = solve(i + solve1(x, y - 1) - solve1(x, y), 0, 0);
    else
      ans = solve(i + solve1(x - 1, y) - solve1(x, y), 0, 0);
    return ans;
  }
  ans = 0;
  if (i) ans = (1LL * i * solve(i - 1, x, y) + ans) % m;
  if (x) {
    long long cnt = solve1(x - 1, y) - solve1(x, y);
    ans = (1LL * cnt * solve(cnt - 1 + i, x - 1, y) + ans) % m;
  }
  if (y) {
    long long cnt = solve1(x, y - 1) - solve1(x, y);
    ans = (1LL * cnt * solve(cnt - 1 + i, x, y - 1) + ans) % m;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  p1[0] = p2[0] = 1;
  for (int i = 1; i <= 19; i++) p1[i] = 2 * p1[i - 1];
  p2[1] = 3;
  memset(dp, -1, sizeof dp);
  cin >> n;
  int x = log2(n), y = 0;
  long long ans = solve(0, x, y);
  if (p1[x - 1] * 3 <= n) ans += solve(0, x - 1, 1);
  ans %= m;
  cout << ans;
}
