Codeforces-150B-Quantity of Strings（思维）

B. Quantity of Strings

time limit per test：2 seconds

memory limit per test：256 megabytes

input：standard input

output：standard output

Just in case somebody missed it: this winter is totally cold in Nvodsk! It is so cold that one gets funny thoughts. For example, let's say there are strings with the length exactly n, based on the alphabet of size m. Any its substring with length equal to k is a palindrome. How many such strings exist? Your task is to find their quantity modulo 1000000007 (10⁹ + 7). Be careful and don't miss a string or two!

Let us remind you that a string is a palindrome if it can be read the same way in either direction, from the left to the right and from the right to the left.

Input

The first and only line contains three integers: n, m and k (1 ≤ n, m, k ≤ 2000).

Output

Print a single integer — the number of strings of the described type modulo 1000000007 (10⁹ + 7).

Examples

Input

1 1 1

Output

1

Input

5 2 4

Output

2

Note

In the first sample only one string is valid: "a" (let's denote the only letter of our alphabet as "a").

In the second sample (if we denote the alphabet letters as "a" and "b") the following strings are valid: "aaaaa" and "bbbbb".

题意：给m个字符，要求用任意的字符组成一个长度为n并且任意长度为k的字都是回文串，问有多少种组合方法；

思路：分以下几种情况：

1、k<n&&k%2==0  此时串中只能有一种字符

2、k==n  （k为奇数，最中间的字符为任意，然后考虑左边或者右边，一边的每个位可以为任意字符） （k位偶数，考虑一边即可）

3、k>n||k==1  每个位可以位任意字符 （k>n是个坑。。。）

4、k为其他情况，（n为奇数，如同ababa这样或者aaaaa，每个字符串只能有一种或两种字符，即m*m） （n为偶数则字符串只能有一种字符）

AC代码：

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<queue>
using namespace std;
int n,m,k;
int main()
{
    while(~scanf("%d%d%d",&n,&m,&k))
    {
        if(k<n&&k%2==0){
            printf("%d\n",m);
            continue;
        }
        long long ans=1;
        if(k==n){
            if(n%2){
                ans=m;
                for(int i=1;i<=k/2;i++)
                    ans=ans*m%1000000007;
            }
            else{
                for(int i=1;i<=k/2;i++)
                    ans=ans*m%1000000007;
            }
        }
        else if(k==1||k>n)
        {
            for(int i=1;i<=n;i++)
                ans=ans*m%1000000007;
        }
        else
        {
            if(!n%2) ans=m;
            else ans=m*m;
        }
        printf("%lld\n",ans);
    }
    return 0;
}