【CodeForces】337C - Quiz（贪心，思维）

题目链接：点击打开题目

C. Quiz

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Manao is taking part in a quiz. The quiz consists of n consecutive questions. A correct answer gives one point to the player. The game also has a counter of consecutive correct answers. When the player answers a question correctly, the number on this counter increases by 1. If the player answers a question incorrectly, the counter is reset, that is, the number on it reduces to 0. If after an answer the counter reaches the number k, then it is reset, and the player's score is doubled. Note that in this case, first 1 point is added to the player's score, and then the total score is doubled. At the beginning of the game, both the player's score and the counter of consecutive correct answers are set to zero.

Manao remembers that he has answered exactly m questions correctly. But he does not remember the order in which the questions came. He's trying to figure out what his minimum score may be. Help him and compute the remainder of the corresponding number after division by 1000000009 (109 + 9).

Input

The single line contains three space-separated integers n, m and k (2 ≤ k ≤ n ≤ 109; 0 ≤ m ≤ n).

Output

Print a single integer — the remainder from division of Manao's minimum possible score in the quiz by 1000000009 (109 + 9).

Examples

input

5 3 2

output

3

input

5 4 2

output

6

Note

Sample 1. Manao answered 3 questions out of 5, and his score would double for each two consecutive correct answers. If Manao had answered the first, third and fifth questions, he would have scored as much as 3 points.

Sample 2. Now Manao answered 4 questions. The minimum possible score is obtained when the only wrong answer is to the question 4.

Also note that you are asked to minimize the score and not the remainder of the score modulo 1000000009. For example, if Manao could obtain either 2000000000 or 2000000020 points, the answer is 2000000000 mod 1000000009, even though 2000000020 mod 1000000009 is a smaller number.

把数分为k段，每一段的最后一个尽量让其错误。

如果不行的话，让连续的k道题都对的情况靠前。然后求连续k*ant道题全对的分数，加上剩余的题数*1就是最后的最小分数了。

代码如下：

#include <cstdio>
#include <stack>
#include <queue>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define LL long long
const LL MOD = 1e9+9;
LL quick_mod(LL a,LL b)
{
	LL mul = 1;
	while (b)
	{
		if (b & 1)
			mul = mul * a % MOD;
		a = a * a % MOD;
		b >>= 1;
	}
	return mul;
}
int main()
{
	LL n,m,k;
	while (~scanf ("%lld %lld %lld",&n,&m,&k))
	{
		LL wa = n - m;
		LL can = n / k;
		if (can <= wa)		//没有翻倍的情况 
			printf ("%lld\n",m % MOD);
		else
		{
			LL ant = can - wa;		//翻倍的次数 
			LL ans = (quick_mod(2,ant) * k - k) * 2 % MOD;
			ans += m - ant * k;		//剩余的题 
			printf ("%lld\n",ans % MOD);
		}
	}
	return 0;
}