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;
long long MOD = 1e9 + 7;
long long dp[2][21][2];
long long remDiv(int n, int cur, int x) { return n / x - cur; }
long long strictlyDiv(int n, int x, int y) { return n / x - n / y; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  int n;
  cin >> n;
  int highest = 31 - __builtin_clz(n);
  if (n == 1) {
    cout << "1\n";
    return 0;
  }
  dp[0][highest][0] = 1;
  if ((1 << (highest - 1)) * 3 <= n) dp[0][highest - 1][1] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = highest; j >= 0; j--) {
      for (int k = 1; k >= 0; k--) {
        int cur = i % 2;
        int prev = 1 - cur;
        int g = (1 << j);
        if (k) g *= 3;
        int same = remDiv(n, i - 1, g);
        if (same < 0) same = 0;
        dp[cur][j][k] = (dp[prev][j][k] * same) % MOD;
        if (j < highest && i > 1) {
          int dec2 = strictlyDiv(n, g, g * 2);
          assert(dec2 >= 0);
          dp[cur][j][k] += (dp[prev][j + 1][k] * dec2) % MOD;
          dp[cur][j][k] %= MOD;
        }
        if (k == 0 && i > 1) {
          int dec3 = strictlyDiv(n, g, g * 3);
          assert(dec3 >= 0);
          dp[cur][j][k] += (dp[prev][j][k + 1] * dec3) % MOD;
          dp[cur][j][k] %= MOD;
        }
      }
    }
  }
  long long ans = dp[n % 2][0][0];
  if (ans < 0) ans += MOD;
  cout << ans << "\n";
}
