C. Two Arrays（组合数学，n个数选择m个数，单调不递减个数，排列组合打表N*N）

题目链接：http://codeforces.com/problemset/problem/1288/C

C. Two Arrays

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given two integers nn and mm. Calculate the number of pairs of arrays (a,b)(a,b) such that:

• the length of both arrays is equal to mm;
• each element of each array is an integer between 11 and nn (inclusive);
• ai≤biai≤bi for any index ii from 11 to mm;
• array aa is sorted in non-descending order;
• array bb is sorted in non-ascending order.

As the result can be very large, you should print it modulo 109+7109+7.

Input

The only line contains two integers nn and mm (1≤n≤10001≤n≤1000, 1≤m≤101≤m≤10).

Output

Print one integer – the number of arrays aa and bb satisfying the conditions described above modulo 109+7109+7.

Examples

input

Copy

2 2

output

Copy

5

input

Copy

10 1

output

Copy

55

input

Copy

723 9

output

Copy

157557417

Note

In the first test there are 55 suitable arrays:

• a=[1,1],b=[2,2]a=[1,1],b=[2,2];
• a=[1,2],b=[2,2]a=[1,2],b=[2,2];
• a=[2,2],b=[2,2]a=[2,2],b=[2,2];
• a=[1,1],b=[2,1]a=[1,1],b=[2,1];
• a=[1,1],b=[1,1]a=[1,1],b=[1,1].

题目大意：

给定一个数n和一个数m，让构建两个数组a和b满足条件，

1.数组中所有元素的取值在1~n之间，a和b数组长度是m。2. a数组是单调不递减的，b数组是单调不递增 3. 任意的位置i，有a_i<=b_i

 问你有多少对这样的数组

思路：

从n个数中任选m个数，这m个数从小到大排列，且可重复选取的方案数为C(n+m-1,m)

 

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<stack>
#include<map>
#include<queue>
using namespace std;
typedef long long LL;
#define sc1(a) scanf("%lld",&a)
#define pf1(a) printf("%lld\n",a)
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
const int INF=1e9+7;
const int maxn=1e3+50;
const int maxm=10+5;
const int maxv=1e6+5;
const int mod=1e9+7;
const int ba=3e5;
LL c[maxn][maxn];
LL N,M;
void Init()
{
    for(int i=1;i<maxn;i++)//
    {
        for(int j=1;j<=i;j++)
        {
            if(j==1) c[i][j]=i;
            else
            {
                c[i][j]=(c[i-1][j]+c[i-1][j-1])%mod;
            }
        }
    }
}
int main()
{
//    freopen("in.txt","r",stdin);
    sc1(N);sc1(M);
    Init();
    LL ans=0,pre=0;
    for(int i=1;i<=N;i++) //从i个元素里选m个元素以i结尾 按非递减排列个数
    {
        ans=(ans+((c[i+M-1][M]-pre+mod)%mod)*(c[N-i+M][M])%mod)%mod;
        pre=c[i+M-1][M];//因为要以i结尾 所以减掉不是以i结尾的
    }
    pf1(ans);
    return 0;
}
/**

*/