C++ :

#pragma GCC optimize ("O2")
#include<cstdio>
#define ll long long
const int N=1e5+5,M=2e5+5,mo=1e9+7;
int rn,n,m,i,j,ans;
int fact[M],inv[M],fv[M];
void init(int n){
	fact[0]=fact[1]=inv[1]=fv[0]=fv[1]=1;
	for (i=2;i<=n;i++){
		fact[i]=(ll)fact[i-1]*i%mo;
		inv[i]=(ll)inv[mo%i]*(mo-mo/i)%mo;
		fv[i]=(ll)fv[i-1]*inv[i]%mo;
	}
}
int cal(int k){
	int n=rn+k-1,m=rn-1;
	return (ll)fact[n]*fv[m]%mo*fv[n-m]%mo;
}
int sum,f[N],dp[N];
void solve(){
	dp[0]=1;f[0]=1;
	for (i=1;i<=n;i++){
		sum+=i;if (sum>m) break;
		for (j=m;j>=sum;j--){
			dp[j]=-dp[j-i];
			if (j>=n+1) dp[j]=(dp[j]+dp[j-n-1])%mo;
		}
		for (;j>=sum-i;j--) dp[j]=0;
		for (j=sum;j<=m;j++){
			dp[j]=(dp[j]+dp[j-i])%mo;
			f[j]=(f[j]+dp[j])%mo;
		}
	}
}
int main(){
	scanf("%d%d",&n,&m);rn=n;
	init(n+m);solve();
	for (i=0;i<=m;i++)
		ans=(ans+(ll)f[i]*cal(m-i))%mo;
	printf("%d",(ans+mo)%mo);
}