cf 403D

/* cf 403D */
/* differences between any pairs should be distinct..so instead of k pairs we can assume k distinct numbers..bi - ai = ci..sum of all ci <= n..why ? coz otherwise we have to take a pair that is greater than 1000... as 1 + 2 +....+ 45 >1000..
so if k > 44 then ans will b zero...

we will recursively iterate all numbers and try to take how many ways are there to divide i into k parts..here i is from 1 to n...but now the most interesting part comes here..we actually can increase the difference between any segments...but sum of all should b n...so stars and bars comes here...
*/

ll f[500];int dp[1001][50][1001];ll ncr[1001][50],ans[1001][50];

ll NCR(int n,int r){
    if(n==r)return 1;
    if(r==1)return n;
    if(n<r)return 0;
    if(ncr[n][r])return ncr[n][r];
    return ncr[n][r] =( NCR(n-1,r-1) + NCR(n-1,r) )%mod;
}


void pre(){
    f[0] = 1;
    for(int i = 1;i<450;i++)f[i] = (f[i-1] * i )%mod;
}

ll stars_and_bars(int n,int k){
    return NCR(n+k-1,k-1);
}

int giveres(int n,int k,int p){
    if(n==0&&k==0)return 1;
    if(n==0||k==0||p>n)return 0;
    int &ret = dp[n][k][p];
    if(ret!=-1)return ret;
    ret = 0;
    ret+=giveres(n-p,k-1,p+1);
    ret %= mod;
    ret+=giveres(n,k,p+1);
    ret %= mod;
    return ret;
}

ll solve(int n,int k){
    if(ans[n][k]!=-1)return ans[n][k];
    ll ret = 0;
    for(int i = n;i>=1;i--){
        ret += ((ll)giveres(i,k,1) * stars_and_bars(n-i,k+1) )%mod;
        ret %= mod;
    }
    return ans[n][k] = (ret * f[k])%mod;
}


int main()
{
    booster()
   // read("input.txt");
    pre();
    mem(dp,-1);
    mem(ans,-1);
    int t;
    cin>>t;
    while(t--){
        int n,k;
        cin>>n>>k;
        if(k>44)cout<<0<<endl;
        else
            cout<<solve(n,k)<<endl;
    }


    return 0;
}