C. Two Arrays

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

可以将题目要求转换为：

1、长度为2*m
2、ci的取值范围为[1,n]
3、序列c非递减

我想了好久终于想明白dp转移方程的道理：

dp[i][j]表示满足以下条件的序列个数：
1、长度为i
2、第一个数为j
3、序列非递减,

那么，dp[i][j]=(dp[i][j + 1] + dp[i - 1][j]) % mod

比如在样例2中，i=2，j=10的时候，如何推出目前的序列数量呢？i=1，j=9的时候，只有{9}这种情况；i=2，j=10的时候，只有{10,10}这种情况

i=1，j=9到i=2，j=9，只需要在i=1，j=9所构成的序列中把序列中任意一个数重复一遍即{9，9}这种情况；

而i=2，j=10到i=2，j=9，只需要在i=2，j=9所构成的序列里，在最前面加上这个数就满足了题目的要求：{9，10}。

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <iomanip>
#include <deque>
#include <bitset>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>
//#include <xfunctional>
#define ll              long long
#define PII             pair<int, int>
#define rep(i,a,b)      for(int  i=a;i<=b;i++)
#define dec(i,a,b)      for(int  i=a;i>=b;i--)
#define pb              push_back
#define mk              make_pair
using namespace std;
int dir1[6][2] = { { 0,1 } ,{ 0,-1 },{ 1,0 },{ -1,0 },{ 1,1 },{ -1,1 } };
int dir2[6][2] = { { 0,1 } ,{ 0,-1 },{ 1,0 },{ -1,0 },{ 1,-1 },{ -1,-1 } };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = 3.14159265358979;
const int mod = 1000000007;
const int N = 1005;
//if(x<0 || x>=r || y<0 || y>=c)

inline ll read()
{
    ll x = 0; bool f = true; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? x : -x;
}
ll gcd(ll m, ll n)
{
    return n == 0 ? m : gcd(n, m % n);
}

const ll mo = 1e9 + 7;
ll dp[22][1100];
int main()
{
    int n, m;
    cin >> n >> m;
    m <<= 1;
    for (int i = 1; i <= n; i++)
        dp[1][i] = 1;
    for (int i = 2; i <= m; i++)
        for (int j = n; j>0; j--)
            dp[i][j] = (dp[i][j + 1] + dp[i - 1][j]) % mo;
    ll ans = 0;
    for (int i = 1; i <= n; i++)
        ans = (ans + dp[m][i]) % mo;
    cout << ans << endl;
    return 0;
}