Problem: <image>

To monitor cryptocurrency exchange rates trader William invented a wonderful device consisting of n lights arranged in a row. The device functions in the following way:

Initially, all lights on William's device are turned off. At the beginning of a new iteration the device randomly, with a uniform distribution, picks a light that is turned off and turns it on, telling William which cryptocurrency he should invest in. After this iteration if any k consecutive lights contain more than one turned on light, then the device finishes working.

William doesn't like uncertainty, so he wants you to calculate the expected value of the number of lights that are turned on in the device after it finishes working.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10). Description of the test cases follows.

The only line for each test case contains two integers n and k (2 ≤ k ≤ n ≤ 10^5), which are the total number of lights and the length of subsegment of lights that are being checked, respectively.

Output

For each test case print the answer, modulo 10^9+7. 

Formally, let M = 10^9+7. It can be shown that the answer can be expressed as an irreducible fraction p/q, where p and q are integers and q not ≡ 0 \pmod{M}. Output the integer equal to p ⋅ q^{-1} mod M. In other words, output such an integer x that 0 ≤ x < M and x ⋅ q ≡ p \pmod{M}.

Example

Input


3
3 2
15 2
40 15


Output


333333338
141946947
329622137

Note

Explanation of the first sample test case:

Let's write out all possible sequences of light toggles, which will make the device complete its operation:

  1. (1, 2) — 2 lights are turned on 
  2. (1, 3, 2) — 3 lights are turned on 
  3. (2, 1) — 2 lights are turned on 
  4. (2, 3) — 2 lights are turned on 
  5. (3, 2) — 2 lights are turned on 
  6. (3, 1, 2) — 3 lights are turned on 



Then the final expected value will be equal to 2/6 + 3/6 + 2/6 + 2/6 + 2/6 + 3/6 = 14/6 = 7/3. 

Then the required output will be 333333338, since 333333338 ⋅ 3 ≡ 7 \pmod{10^9+7}.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.math.BigInteger;

/**
 __                  __
 ( _)                ( _)
 / / \\              / /\_\_
 / /   \\            / / | \ \
 / /     \\          / /  |\ \ \
 /  /   ,  \ ,       / /   /|  \ \
 /  /    |\_ /|      / /   / \   \_\
 /  /  |\/ _ '_| \   / /   /   \    \\
 |  /   |/  0 \0\    / |    |    \    \\
 |    |\|      \_\_ /  /    |     \    \\
 |  | |/    \.\ o\o)  /      \     |    \\
 \    |     /\\`v-v  /        |    |     \\
 | \/    /_| \\_|  /         |    | \    \\
 | |    /__/_ `-` /   _____  |    |  \    \\
 \|    [__]  \_/  |_________  \   |   \    ()
 /    [___] (    \         \  |\ |   |   //
 |    [___]                  |\| \|   /  |/
 /|    [____]                  \  |/\ / / ||
 (  \   [____ /     ) _\      \  \    \| | ||
 \  \  [_____|    / /     __/    \   / / //
 |   \ [_____/   / /        \    |   \/ //
 |   /  '----|   /=\____   _/    |   / //
 __ /  /        |  /   ___/  _/\    \  | ||
 (/-(/-\)       /   \  (/\/\)/  |    /  | /
 (/\/\)           /   /   //
 _________/   /    /
 \____________/    (


 @author NTUDragons-Reborn
 */

public class Solution{
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Task solver = new Task();
        solver.solve(in, out);
        out.close();
    }
    // main solver
    static class Task{

        double eps= 0.00000001;
        static final int MAXN = 100005;
        static final int MOD= 1000000007;

        // stores smallest prime factor for every number
        static int spf[] = new int[MAXN];

        Map<Integer,Set<Integer>> dp= new HashMap<>();

        // Calculating SPF (Smallest Prime Factor) for every
        // number till MAXN.
        // Time Complexity : O(nloglogn)
        public void sieve()
        {
            spf[1] = 1;
            for (int i=2; i<MAXN; i++)

                // marking smallest prime factor for every
                // number to be itself.
                spf[i] = i;

            // separately marking spf for every even
            // number as 2
            for (int i=4; i<MAXN; i+=2)
                spf[i] = 2;

            for (int i=3; i*i<MAXN; i++)
            {
                // checking if i is prime
                if (spf[i] == i)
                {
                    // marking SPF for all numbers divisible by i
                    for (int j=i*i; j<MAXN; j+=i)

                        // marking spf[j] if it is not
                        // previously marked
                        if (spf[j]==j)
                            spf[j] = i;
                }
            }
        }

        // A O(log n) function returning primefactorization
        // by dividing by smallest prime factor at every step
        public Set<Integer> getFactorization(int x)
        {
            if(dp.containsKey(x)) return dp.get(x);
            Set<Integer> ret = new HashSet<>();
            while (x != 1)
            {
                if(spf[x]!=2) ret.add(spf[x]);
                x = x / spf[x];
            }
            dp.put(x,ret);
            return ret;
        }
        // function to find first index >= x
        public int lowerIndex(List<Integer> arr, int n, int x)
        {
            int l = 0, h = n - 1;
            while (l <= h)
            {
                int mid = (l + h) / 2;
                if (arr.get(mid) >= x)
                    h = mid - 1;
                else
                    l = mid + 1;
            }
            return l;
        }

        // function to find last index <= y
        public int upperIndex(List<Integer> arr, int n, int y)
        {
            int l = 0, h = n - 1;
            while (l <= h)
            {
                int mid = (l + h) / 2;
                if (arr.get(mid) <= y)
                    l = mid + 1;
                else
                    h = mid - 1;
            }
            return h;
        }

        // function to count elements within given range
        public int countInRange(List<Integer> arr, int n, int x, int y)
        {
            // initialize result
            int count = 0;
            count = upperIndex(arr, n, y) -
                    lowerIndex(arr, n, x) + 1;
            return count;
        }

        public int add(int a, int b){
            a+=b;
            if(a>=MOD) a-=MOD;
            else if(a<0) a+=MOD;
            return a;
        }

        public int mul(int a, int b){
            long res= (long)a*(long)b;
            return (int)(res%MOD);
        }

        public int power(int a, int b) {
            int ans=1;
            while(b>0){
                if((b&1)!=0) ans= mul(ans,a);
                b>>=1;
                a= mul(a,a);
            }
            return ans;
        }

        int[] fact= new int[MAXN];
        int[] inv= new int[MAXN];

        public int Ckn(int n, int k){
            if(k<0 || n<0) return 0;
            return mul(mul(fact[n],inv[k]),inv[n-k]);
        }

        public int inverse(int a){
            return power(a,MOD-2);
        }

        public void preprocess() {
            fact[0]=1;
            for(int i=1;i<MAXN;i++) fact[i]= mul(fact[i-1],i);
            inv[MAXN-1]= inverse(fact[MAXN-1]);
            for(int i=MAXN-2;i>=0;i--){
                inv[i]= mul(inv[i+1],i+1);
            }
        }
        
        
        public void solve(InputReader in, PrintWriter out) {
            /*
            E(x) : expected number of lights on immediately before terminating
            P(x>=d) : probability of turning on d-1 elements such that each adjacent elements are k aprt
            E(x)= P(x=1)
                + P(x=2) + P(x=2)
                + P(x=3) + P(X=3) + P(x=3)
                ..
                + P(X=n) + P(x=n) + ...    + P(X=n)
            
            => E(X)= (P(x=1)+P(x=2)+...+P(x=n))+ (P(x=2)+P(x=3)+..+P(x=n))+...
            =P(x>=1)+P(x>=2)+...+P(x>=n)

            P(x>=p): 
            _ First , choose d lights from n lights to turn on
            +Probability to choost first light : p/n
            +Probability to choost second light : p-1/n-1
            ...
            --> Probability to choose p lights to turn on : p!/((n-1)(n-2)...(n-p+1))
            
            _How many ways to choose such p: between such p turned on, there must be at least k-1 turned off lights to 
            gurantee that the system not terminated yet.
            + First put k-1 turned off lights between p turned on light 
            --> Number of lights left: n-p-(k-1)*(p-1). 
            +Put those lights into p+1 places among p on-lights after inserting k-1 lights between
            --> Based on star and bar theorem, nunmber of ways : C(n-(k-1)*(p-1),p)

            => Expected number of turned on lights one way before terminating: Zigma(p:1->n)(C(n-(k-1)*(p-1),p)*(p!/((n-1)(n-2)...(n-p+1))))
            => FInal answer + 1: to terminate process 
            */
            
            preprocess();
            int t= in.nextInt();
            while(t-->0){
                int n= in.nextInt(), k= in.nextInt();
                int res=1;
                for(int i=1;i<=n;i++){
                    if(n-i-(k-1)*(i-1)<0) break;
                    res= add(res,mul( mul( mul( Ckn(n-(k-1)*(i-1),i) , fact[i]) , inv[n]), fact[n-i]));
                }
                out.println(res);
            }
        }
    

        // public static class compareL implements Comparator<Tuple>{
        //     @Override
        //     public int compare(Tuple t1, Tuple t2) {
        //         return t2.l - t1.l;
        //     }
        // }
        
        public int _gcd(int a,  int  b)
        {

            if(b == 0) {
                return a;
            }
            else {
                return _gcd(b, a % b);
            }
        }

    }

    static class Venice{
        public Map<Long,Long> m= new HashMap<>();
        public long base=0;
        public long totalValue=0;
        private int M= 1000000007;

        private long addMod(long a, long b){
            a+=b;
            if(a>=M) a-=M;
            return a;
        }

        public void reset(){
            m= new HashMap<>();
            base=0;
            totalValue=0;
        }
        public void update(long add){
            base= base+ add;
        }
        public void add(long key, long val){
            long newKey= key-base;
            m.put(newKey, addMod(m.getOrDefault(newKey,(long)0),val));
        }
    }


    static class Tuple implements Comparable<Tuple>{
        int x, y, z;
        public Tuple(int x, int y, int z){
            this.x= x;
            this.y= y;
            this.z=z;
        }
        @Override
        public int compareTo(Tuple o){
            return this.x-o.x;
        }
    }

    static class Pair implements Comparable<Pair>{
        public int x;
        public int y;
        public Pair(int x, int y){
            this.x= x;
            this.y= y;
        }

        @Override
        public int compareTo(Pair o) {
            return this.x-o.x;
        }
    }
    // fast input reader class;
    static class InputReader {
        BufferedReader br;
        StringTokenizer st;

        public InputReader(InputStream stream) {
            br = new BufferedReader(new InputStreamReader(stream));
        }

        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                String line = null;
                try {
                    line = br.readLine();
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
                if (line == null) {
                    return null;
                }
                st = new StringTokenizer(line);
            }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(nextToken());
        }
        public double nextDouble(){
            return Double.parseDouble(nextToken());
        }
        public long nextLong(){
            return Long.parseLong(nextToken());
        }
    }
}