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>
long long dx[8] = {0, 1, 0, -1, 1, 1, -1, -1};
long long dy[8] = {1, 0, -1, 0, -1, 1, 1, -1};
using namespace std;
class pa3 {
 public:
  long long x;
  long long y, z;
  pa3(long long x = 0, long long y = 0, long long z = 0) : x(x), y(y), z(z) {}
  bool operator<(const pa3 &p) const {
    if (x != p.x) return x < p.x;
    if (y != p.y) return y < p.y;
    return z < p.z;
  }
  bool operator>(const pa3 &p) const {
    if (x != p.x) return x > p.x;
    if (y != p.y) return y > p.y;
    return z > p.z;
  }
  bool operator==(const pa3 &p) const {
    return x == p.x && y == p.y && z == p.z;
  }
  bool operator!=(const pa3 &p) const {
    return !(x == p.x && y == p.y && z == p.z);
  }
};
class pa4 {
 public:
  long long x;
  long long y, z, w;
  pa4(long long x = 0, long long y = 0, long long z = 0, long long w = 0)
      : x(x), y(y), z(z), w(w) {}
  bool operator<(const pa4 &p) const {
    if (x != p.x) return x < p.x;
    if (y != p.y) return y < p.y;
    if (z != p.z) return z < p.z;
    return w < p.w;
  }
  bool operator>(const pa4 &p) const {
    if (x != p.x) return x > p.x;
    if (y != p.y) return y > p.y;
    if (z != p.z) return z > p.z;
    return w > p.w;
  }
  bool operator==(const pa4 &p) const {
    return x == p.x && y == p.y && z == p.z && w == p.w;
  }
};
class pa2 {
 public:
  long long x, y;
  pa2(long long x = 0, long long y = 0) : x(x), y(y) {}
  pa2 operator+(pa2 p) { return pa2(x + p.x, y + p.y); }
  pa2 operator-(pa2 p) { return pa2(x - p.x, y - p.y); }
  bool operator<(const pa2 &p) const { return y != p.y ? y < p.y : x < p.x; }
  bool operator>(const pa2 &p) const { return x != p.x ? x < p.x : y < p.y; }
  bool operator==(const pa2 &p) const {
    return abs(x - p.x) == 0 && abs(y - p.y) == 0;
  }
  bool operator!=(const pa2 &p) const {
    return !(abs(x - p.x) == 0 && abs(y - p.y) == 0);
  }
};
string itos(long long i) {
  ostringstream s;
  s << i;
  return s.str();
}
long long gcd(long long v, long long b) {
  if (v > b) return gcd(b, v);
  if (v == b) return b;
  if (b % v == 0) return v;
  return gcd(v, b % v);
}
long long mod;
long long extgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long d = extgcd(b, a % b, y, x);
  y -= a / b * x;
  return d;
}
pair<long long, long long> operator+(const pair<long long, long long> &l,
                                     const pair<long long, long long> &r) {
  return {l.first + r.first, l.second + r.second};
}
pair<long long, long long> operator-(const pair<long long, long long> &l,
                                     const pair<long long, long long> &r) {
  return {l.first - r.first, l.second - r.second};
}
long long pr[2200010];
long long inv[2200010];
long long beki(long long wa, long long rr, long long warukazu) {
  if (rr == 0) return 1 % warukazu;
  if (rr == 1) return wa % warukazu;
  wa %= warukazu;
  if (rr % 2 == 1)
    return ((long long)beki(wa, rr - 1, warukazu) * (long long)wa) % warukazu;
  long long zx = beki(wa, rr / 2, warukazu);
  return (zx * zx) % warukazu;
}
long long comb(long long nn, long long rr) {
  if (rr < 0 || rr > nn || nn < 0) return 0;
  long long r = pr[nn] * inv[rr];
  r %= mod;
  r *= inv[nn - rr];
  r %= mod;
  return r;
}
void gya(long long ert) {
  pr[0] = 1;
  for (long long i = 1; i <= ert; i++) {
    pr[i] = (pr[i - 1] * i) % mod;
  }
  inv[ert] = beki(pr[ert], mod - 2, mod);
  for (long long i = ert - 1; i >= 0; i--) {
    inv[i] = inv[i + 1] * (i + 1) % mod;
  }
}
long long n;
long long solve(vector<long long> ve) {
  long long k = ve.size();
  for (long long i = 1; i < k; i++) ve[i] *= ve[i - 1];
  vector<long long> kosu;
  kosu.push_back(n - n / ve[0]);
  for (long long i = 0; i < k - 1; i++) {
    kosu.push_back(n / ve[i] - n / ve[i + 1]);
  }
  kosu.push_back(n / ve[k - 1]);
  long long ans = 1;
  long long ru = 0;
  for (long long i = 0; i <= k; i++) {
    ans *= comb(ru + kosu[i] - 1, ru);
    ans %= 1000000007;
    ans *= pr[kosu[i]];
    ans %= 1000000007;
    ru += kosu[i];
  }
  return ans;
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  mod = 1000000007;
  gya(2000200);
  long long ans = 0;
  cin >> n;
  long long g = 1;
  vector<long long> ve;
  while (1) {
    if (g * 2 > n) break;
    ve.push_back(2);
    g *= 2;
  }
  ans += solve(ve);
  if (g / 2 * 3 <= n) {
    for (auto &v : ve) {
      v = 3;
      ans += solve(ve);
      v = 2;
    }
  }
  cout << ans % 1000000007 << endl;
  return 0;
}
