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 CPP):

Solution:
#include<bits/stdc++.h>
#define reg register 
typedef long long ll;
using namespace std;
inline int qr(){
	int x=0,f=0;char ch=0;
	while(!isdigit(ch)){f|=ch=='-';ch=getchar();}
	while(isdigit(ch)){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
	return f?-x:x;
} 
const int N=1e5+100,mod=1e9+7;
int n,K;
ll qkpow(ll a,ll b){
	ll res=1;
	while(b){
		if(b&1) res=res*a%mod;
		a=a*a%mod,b>>=1;
	}
	return res;
}
ll jie[N],ni[N];
ll C(int a,int b){
	if(b>a||b<0||a<0) return 0;
	return jie[a]*ni[a-b]%mod*ni[b]%mod;
}
int main(){
	int T=qr();		
	jie[0]=ni[0]=1;
	for(reg int i=1;i<=100000;i++){
		jie[i]=jie[i-1]*i%mod;		
	}
	ni[100000]=qkpow(jie[100000],mod-2);
	for(reg int i=99999;i;i--){
		ni[i]=ni[i+1]*(i+1)%mod;
	}
	while(T--){
		n=qr(),K=qr();
		ll ans=1;
		for(reg int i=1;(i-1)*(K-1)<=n;i++){//已经选择i个,下一次出现连续的方案数量,每次计算一个增量,全部累加起来就是最终的贡献 
			ans=(ans+C(n-(ll)(i-1)*(K-1),i)*jie[i]%mod*jie[n-i]%mod*ni[n]%mod)%mod;
		} 
		printf("%lld\n",ans);		
	}
	return 0;
}
/*
较为感性的理解: 
关于答案的统计,每次计算的所有情况都会在分子上贡献1,首先是选择一个,其次是选择2个,选择2个一定是由选择一个拓展来的
所以说考虑选择2个的方案的时候其中1盏灯的贡献已经计算过了，所以仍然只加上方案的数量即可,最终直接+1是因为在这所有的下一步就会终结的情况中,再+1就能够结束 

*/