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 + 5;
const long long MOD = 1e9 + 7;
int n;
long long DP[2][25][2];
inline int Read() {
  int x = 0, f = 0;
  char c = getchar();
  while (!isdigit(c)) {
    f |= c == '-';
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return f ? -x : x;
}
inline void Write(const long long &x) {
  if (x < 0) {
    putchar('-');
    Write(-x);
    return;
  }
  if (x > 9) {
    Write(x / 10);
  }
  putchar((x % 10) ^ 48);
  return;
}
inline long long Solve(const int &x, const int &y) {
  return (1LL << x) * (3LL - y);
}
int main() {
  n = Read();
  register int i, j, Log;
  for (Log = -1; (1 << (Log + 1)) <= n; ++Log)
    ;
  if ((1 << (Log - 1)) * 3 <= n) {
    DP[1][Log - 1][1] = 1;
  }
  DP[1][Log][0] = 1;
  for (i = 2; i <= n; ++i) {
    for (j = 0; j <= Log; ++j) {
      DP[i & 1][j][0] =
          DP[(i - 1) & 1][j][0] * (n / (1LL << j) - i + 1) +
          DP[(i - 1) & 1][j + 1][0] * 1LL *
              (n / Solve(j, 2) - n / Solve(j + 1, 2)) +
          DP[(i - 1) & 1][j][1] * 1LL * (n / Solve(j, 2) - n / Solve(j, 0));
      DP[i & 1][j][1] = DP[(i - 1) & 1][j][1] * (n / ((1LL << j) * 3) - i + 1) +
                        DP[(i - 1) & 1][j + 1][1] * 1LL *
                            (n / Solve(j, 0) - n / Solve(j + 1, 0));
      DP[i & 1][j][0] %= MOD, DP[i & 1][j][1] %= MOD;
    }
  }
  Write(DP[n & 1][0][0]), putchar('\n');
  return 0;
}
