Yesterday, my teacher taught us about math: +, -, *, /, GCD, LCM... As you know, LCM (Least common multiple) of two positive numbers can be solved easily because of a * b = GCD (a, b) * LCM (a, b).

In class, I raised a new idea: "how to calculate the LCM of K numbers". It‘s also an easy problem indeed, which only cost me 1 minute to solve it. I raised my hand and told teacher about my outstanding algorithm. Teacher just smiled and smiled...

After class, my teacher gave me a new problem and he wanted me solve it in 1 minute, too. If we know three parameters N, M, K, and two equations:

1. SUM (A₁, A₂, ..., A_i, A_i+1,..., A_K) = N 
2. LCM (A₁, A₂, ..., A_i, A_i+1,..., A_K) = M

Can you calculate how many kinds of solutions are there for A_i (A_i are all positive numbers). I began to roll cold sweat but teacher just smiled and smiled.

Can you solve this problem in 1 minute?

Input

There are multiple test cases.

Each test case contains three integers N, M, K. (1 ≤ N, M ≤ 1,000, 1 ≤ K ≤ 100)

Output

For each test case, output an integer indicating the number of solution modulo 1,000,000,007(1e9 + 7).

You can get more details in the sample and hint below.

Sample Input

4 2 2
3 2 2

Sample Output

1
2

题意：

给出n,m,k,问k个数的和为n。最小公倍数为m的情况有几种

思路：

由于最小公倍数为m，能够知道这些数必定是m的因子，那么我们仅仅须要选出这全部的因子，拿这些因子来背包就能够了

dp[i][j][k]表示放了i个数，和为j。公倍数为k的情况有几种

可是又问题。首先的问题内存，直接存明显爆内存。那么我们须要优化

1.由于我如今放第i个数，必定是依据放好的i-1个数来计算的。我们仅仅须要用滚动数组来解决就可以

2.对于公倍数，必定不能超过m，而我全部这些m的因子中的数字，不管选哪些，选多少，他们的最小公倍数依旧是这些因子之中的，那么我们能够进行离散化

解决好了之后就是全然背包的问题了

#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <vector>
#include <math.h>
using namespace std;

const int mod = 1e9+7;

int dp[2][1005][105];
int a[1005],len,pos[1005];
int n,m,k;
int hash[1005][1005];

int gcd(int a,int b)
{
    return b==0?
a:gcd(b,a%b);
}
int lcm(int a,int b)
{
    return a/gcd(a,b)*b;
}

int main()
{
    int i,j,x,y;
    for(i = 1; i<=1000; i++)//预处理最小公倍数
    {
        for(j = 1; j<=1000; j++)
            hash[i][j] = lcm(i,j);
    }
    while(~scanf("%d%d%d",&n,&m,&k))
    {
        len = 0;
        memset(pos,-1,sizeof(pos));
        for(i = 1; i<=m; i++)
        {
            if(m%i==0)
            {
                a[len] = i;
                pos[i] = len++;//离散化
            }
        }
        memset(dp[0],-1,sizeof(dp[0]));
        dp[0][0][0] = 1;
        for(i = 1; i<=k; i++)
        {
            memset(dp[i%2],-1,sizeof(dp[i%2]));
            for(j = i-1; j<=n; j++)//由于最小必定放1,而我前面已经放了i-1个数了。前面的和最少必定是i-1
            {
                for(x = 0; x<len; x++)//枚举前面数字的公倍数
                {
                    if(dp[(i+1)%2][j][x]==-1)
                        continue;
                    for(y = 0; y<len && (a[y]+j)<=n; y++)//枚举这一位放哪些数
                    {
                        int r = hash[a[y]][a[x]];
                        int s = j+a[y];
                        if(pos[r]!=-1 && r<=m)
                        {
                            r = pos[r];
                            if(dp[i%2][s][r] == -1) dp[i%2][s][r] = 0;
                            dp[i%2][s][r]+=dp[(i+1)%2][j][x];
                            dp[i%2][s][r]%=mod;
                        }
                    }
                }
            }
        }
        if(dp[k%2][n][pos[m]]==-1)
            printf("0\n");
        else
            printf("%d\n",dp[k%2][n][pos[m]]);
    }

    return 0;
}