Ideas ：

First, recursive , Find out
$f_{i,j}=f_{i-1,j}+f_{i,j-1}$
among i>=j
So we found i==j Time is Cartland number , Then we can push the inner layer step by step
Finally, the simplified formula is $\frac{(n+1-m)\ast (m+n)!}{(m!\ast (n+1)!)}$
（ I'm going to melt myself , A little bit of a problem ）

$code$

#include<iostream>
#include<cstdio>

using namespace std;

int n, m, maxx;
int a[10010], b[10010], c[10001];

void fenj(int x)
{
    
	int j=2;
	while(x!=1)
	{
    
		while(x%j==0)
		{
    
			x/=j;
			a[j]++;
		}
		j++;
	}
	if(x!=1)
		a[x]++;
}

void fenk(int x)
{
    
	int j=2;
	while(x!=1)
	{
    
		while(x%j==0)
		{
    
			x/=j;
			b[j]++;
		}
		j++;
	}
	if(x!=1)
		b[x]++;
}

void cheng(int x)
{
    
	int g=0;
	for(int i=10000; i>=1; i--)
	{
    
		c[i]=c[i]*x+g;
		g=c[i]/10;
		c[i]%=10;
	}
}

void write_()
{
    
	int i=1;
	while(c[i]==0) i++;
	while(i<=10000) printf("%d", c[i]), i++;
}

int main()
{
    
	scanf("%d%d", &n, &m);
	fenj(n+1-m);
	for(int i=1; i<=n+m; i++)
		fenj(i);
	for(int i=1; i<=m; i++)
		fenk(i);
	for(int i=1; i<=n+1; i++)
		fenk(i);
	for(int i=1; i<=10000; i++)
		a[i]-=b[i];
	for(int i=10000; i>=1; i--)
		if(a[i])
		{
    
			maxx=i;
			break;
		}
	c[10000]=1;
	for(int i=1; i<=maxx; i++)
		while(a[i])	a[i]--, cheng(i);
	write_();
	return 0;
}