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 MAX = 1e6 + 5;
const int MOD = 1e9 + 7;
int n, pw[22], dp[MAX][22][2];
int get_g(int e2, int e3) {
  int g_ = pw[e2];
  if (e3 > 0) g_ *= 3;
  return g_;
}
int F(int i, int e2, int e3) {
  if (i == n + 1) {
    return e2 == 0 && e3 == 0;
  }
  int &ans = dp[i][e2][e3];
  if (ans != -1) return ans;
  ans = 0;
  int g_ = get_g(e2, e3);
  for (int x = 0; x < 2; x++) {
    for (int y = 0; y < 2; y++) {
      if (e2 - x < 0 || e3 - y < 0) continue;
      if (x + y < 2) {
        int g = get_g(e2 - x, e3 - y);
        int value = (n / g) - (n / g_);
        if (g == g_) value = (n / g) - (i - 1);
        value = max(0, value);
        int v = 1LL * value * F(i + 1, e2 - x, e3 - y) % MOD;
        ans = (ans + v) % MOD;
      }
    }
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  memset(dp, -1, sizeof(dp));
  pw[0] = 1;
  for (int i = 1; i <= 20; i++) pw[i] = pw[i - 1] * 2;
  int k = log2(n);
  int ans = F(2, k, 0);
  if (3 * (1 << (k - 1)) <= n) ans = (ans + F(2, k - 1, 1)) % MOD;
  printf("%d\n", ans);
  return 0;
}
