斐波那契數列的定義如下： F(0) = 0 F(1) = 1 F(n) = F(n - 1) + F(n - 2) (n >= 2) (1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, ...) 給出n，求F(n)，由於結果很大，輸出F(n) % 1000000009的結果即可。 Input

輸入1個數n(1 <= n <= 10^18)。

Output

輸出F(n) % 1000000009的結果。

Input示例

11

Output示例

89

#include <iostream>
#include <algorithm>
#include <cmath>
#define MOD 1000000009
#define N 2
using namespace std;

struct Matrix 
{
	long long v[N][N];
};

Matrix matrix_mul(Matrix A, Matrix B, long long m)
{
	Matrix ans;

	for (int i = 0; i < N; i++)
	{
		for (int j = 0; j < N; j++)
		{
			ans.v[i][j] = 0;
			for (int k = 0; k < N; k ++)
			{
				ans.v[i][j] += (A.v[i][k] * B.v[k][j]) % m;
			}
			ans.v[i][j] %= m;
		}
	}
	return ans;
}

Matrix matrix_pow(Matrix C, long long n, long long m)
{
	Matrix ans = {1, 0, 0, 1};
	while (n)
	{
		if (n & 1)
			ans = matrix_mul(ans, C, m);
		C = matrix_mul(C, C, m);
		n >>= 1;
	}
	return ans;
}

int main()
{
	long long n;
	Matrix temp1 = {1, 1, 1, 0}, temp2 = {1, 0, 1, 0}; // temp2{f[2],0,f[1],0}!!!!!   temp1[a,b,1,0] ---> f[n]=a*f[n-1]+b*f[n-2]
	while (cin >> n)
	{
		if (n < 2)
		{
			cout << 1 << endl;
			continue;
		}
		Matrix res = matrix_pow(temp1, n - 2, MOD);
		res = matrix_mul(res, temp2, MOD);
		cout << res.v[0][0] << endl;
	}
	return 0;
}