337C

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 ofn 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 numberk, 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 exactlym 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 integersn, m andk (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 by1000000009(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 modulo1000000009. For example, if Manao could obtain either2000000000 or 2000000020 points, the answer is2000000000 mod 1000000009, even though2000000020 mod 1000000009 is a smaller number.

题意：

         给你  n 个问题，其中有m  个你知道答案，当你连续答对 k 个问题时，你的分数先+1 ，然后乘以2；求最小的分数。

思路：很容易想到要先考虑 n-m  是否大于  n/k ，大于的情况很简单，对于小于的情况： 找到多余的个数（对于每隔k个就出现一个答错这种排列多余的个数），然后 找一下可以发现一个简单的规律。

#include<bits/stdc++.h>#define MOD 1000000009using namespace std;typedef long long ll;ll Pow(ll a,int b){    ll sum=1;    while(b)    {        if(b&1)            sum=sum*a%MOD;        a=a*a%MOD;        b>>=1;    }    return sum%MOD;}int main(){    ll n,m,k,t;cin>>n>>m>>k;t=n-m;int tw=n/k;if (t>=tw)        cout<<m%MOD<<endl;else{        t=tw-t;        ll A=((4*Pow(2,t-1))%MOD-2+MOD)%MOD;        ll ans=(A*k%MOD+m-k*t)%MOD;        cout<<ans%MOD<<endl;}return 0;}