Educational Codeforces Round 80 (Rated for Div. 2) C. Two Arrays #数位DP#

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

2 2

output

5

input

10 1

output

55

input

723 9

output

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].

Solution

#include 
using namespace std;

typedef long long ll;
const int maxn = 1e3 + 10;
const int mod = 1e9 + 7;
int dp[11][maxn];

inline const int read()
{
    int x = 0, f = 1; char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = (x << 3) + (x << 1) + ch - '0'; ch = getchar(); }
    return x * f;
}

int main()
{
    int n = read(), m = read(), res = 0;
    for (int i = 1; i <= n; i++) dp[1][i] = 1;
    for (int i = 1; i <= m; i++)
        for (int j = 1; j <= n; j++)
            for (int k = 1; k <= j; k++)
                dp[i][j] = (dp[i][j] + dp[i - 1][k]) % mod;
    for (int i = 1; i <= n; i++)
    {
        ll sum = 0;
        for (int j = 1; j <= n - i + 1; j++)
            sum = (sum + dp[m][j]) % mod;
        res = (res + sum * dp[m][i] % mod) % mod;
    }
    printf("%d\n", res);
    return 0;
}