#include <bits/stdc++.h>
#define int long long
using namespace std;
using ll = long long; 
const int N = 14000005; 
const int mod = 1000000007;

int n;
int fac[N], ifac[N];
int dp[16][65537];

int qpow(int a, int b = mod - 2) {
	int res = 1;
	for (; b; b >>= 1) {
		if(b & 1) res = 1ll * res * a % mod;
		a = 1ll * a * a % mod;
	} 
	return res;
}

int C(int n, int m) {
	if(m > n) return 0;
	return 1ll * fac[n] * ifac[m] % mod * ifac[n - m] % mod; 
}

signed main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin >> n;
	fac[0] = 1;
	for (int i = 1; i < N; i++) fac[i] = 1ll * i * fac[i - 1] % mod;
	ifac[N - 1] = qpow(fac[N - 1]);
	for (int i = N - 1; i; i--) ifac[i - 1] = 1ll * ifac[i] * i % mod; 

	dp[0][1] = 1;
	for (int i = 1; i <= 15; i++) {
		int vac = qpow(3, i - 1);
		int v = qpow(3, i) - qpow(3, i - 1);
		int t1 = qpow(C(n - vac, v));
		if(n - vac <= v) {
			for (int j = 1; j <= (1 << i - 1); j++) 
				dp[i][0] = (dp[i][0] + dp[i - 1][j]) % mod;
			break;
		}
		
//		for (int j = 0; j <= (1 << i); j++) { // 感染的人数 
//			for (int k = max(1ll, (j + 1) >> 1); k <= (1 << i - 1); k++) { // 
//				int l = 2 * k - j; // 感染者中选择接种的数量 
//				dp[i][j] = (dp[i][j] + 1ll * dp[i - 1][k] * t1 % mod * C(k * 2, l) % mod * C(n - k * 2 - vac, v - l) % mod) % mod;
//				assert(k <= n);
//			}
//		}

		for (int j = 1; j <= (1 << i - 1); j++) if(dp[i - 1][j]) {
			for (int k = 0; k <= j * 2; k++) {
				int l = 2 * j - k;
				dp[i][k] = (dp[i][k] + 1ll * dp[i - 1][j] * t1 % mod * C(j * 2, l) % mod * C(n - vac - j * 2, v - l) % mod) % mod;
			}
 		}
	}
	
	int ans = 0;
	for (int i = 1; i <= 15; i++) ans = (ans + 1ll * dp[i][0] * i) % mod;
	cout << ans << "\n";
}