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 maxn = 1e6;
int n, step2[31], step3[31], max_dist;
int dp[maxn + 10][21][2];
long long f(int x, int y) { return (long long)n / (step2[x] * step3[y]); }
int main() {
  scanf("%d", &n);
  int maxp2 = log2(n);
  step2[0] = 1;
  step3[0] = 1;
  for (int i = 1; i <= 30; i++) {
    step2[i] = step2[i - 1] * 2;
    step3[i] = step3[i - 1] * 3;
  }
  max_dist = min(maxp2 + 1, n);
  dp[1][maxp2][0] = 1;
  if ((step2[maxp2] / 2) * 3 <= n) dp[1][maxp2 - 1][1] = 1;
  for (int i = 1; i <= n - 1; i++) {
    for (int j = 0; j <= maxp2; j++) {
      for (int k = 0; k <= 1; k++) {
        if (f(j, k) - i > 0)
          dp[i + 1][j][k] =
              ((long long)dp[i + 1][j][k] +
               (long long)dp[i][j][k] * (f(j, k) - (long long)i)) %
              1000000007;
        if (j > 0)
          dp[i + 1][j - 1][k] =
              ((long long)dp[i + 1][j - 1][k] +
               (long long)dp[i][j][k] * (f(j - 1, k) - f(j, k))) %
              1000000007;
        if (k > 0)
          dp[i + 1][j][k - 1] =
              ((long long)dp[i + 1][j][k - 1] +
               (long long)dp[i][j][k] * (f(j, k - 1) - f(j, k))) %
              1000000007;
      }
    }
  }
  printf("%d\n", dp[n][0][0]);
  return 0;
}
