Let's call an integer array a_1, a_2, ..., a_n good if a_i ≠ i for each i.

Let F(a) be the number of pairs (i, j) (1 ≤ i < j ≤ n) such that a_i + a_j = i + j.

Let's say that an array a_1, a_2, ..., a_n is excellent if: 

  * a is good; 
  * l ≤ a_i ≤ r for each i; 
  * F(a) is the maximum possible among all good arrays of size n. 



Given n, l and r, calculate the number of excellent arrays modulo 10^9 + 7.

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. 

The first and only line of each test case contains three integers n, l, and r (2 ≤ n ≤ 2 ⋅ 10^5; -10^9 ≤ l ≤ 1; n ≤ r ≤ 10^9).

It's guaranteed that the sum of n doesn't exceed 2 ⋅ 10^5.

Output

For each test case, print the number of excellent arrays modulo 10^9 + 7.

Example

Input


4
3 0 3
4 -3 5
42 -33 55
69 -42 146


Output


4
10
143922563
698570404

Note

In the first test case, it can be proven that the maximum F(a) among all good arrays a is equal to 2. The excellent arrays are: 

  1. [2, 1, 2]; 
  2. [0, 3, 2]; 
  3. [2, 3, 2]; 
  4. [3, 0, 1]. 
#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
const long long int MOD=1000000007;
int t,n,l,r;
long long ans,ji[200005],ni[200005];
long long kuai(int pos){
	if(pos==1)return 1;
	if(pos==0)return 0;
	int x=MOD-2;
	long long ret=1,m=pos;
	while(x){
		if(x&1)ret=ret*m%MOD;
		m=m*m%MOD;
		x/=2;
	}
	return ret;
}
long long C(int b,int a){
	if(a==b)return 1;
	if(b==0)return 1;
	long long ret=1;
	ret=ret*ni[b]%MOD*ni[a-b]%MOD*ji[a]%MOD;
	return ret;
}
int main(){
	scanf("%d",&t);
	ji[0]=1;ni[0]=1;
	for(int i=1;i<=200000;i++){
		ji[i]=ji[i-1]*i%MOD;
		ni[i]=ni[i-1]*kuai(i)%MOD;
	}
	while(t--){
		ans=0;
		scanf("%d%d%d",&n,&l,&r);
		long long k=min(1-l,r-n);
		if(n%2){
			ans=(k*C(n/2,n)%MOD+k*C(n/2+1,n)%MOD)%MOD;
			//printf("%lld!!!\n",ans);
			int b=n/2,a=n;
			for(;;){
				k++;
				if(k>1-l){
					b--;
					a--;
				}
				if(k>r-n)a--;//printf("%lld %d %d@\n",k,b,a);
				if(a<0||b<-1||a<b)break;
				ans=(ans+C(b+1,a))%MOD;
				if(b>=0)ans=(ans+C(b,a))%MOD;
			}
		}
		else{
			ans=k*C(n/2,n)%MOD;
			int b=n/2,a=n;
			for(;;){
				k++;
				if(k>1-l){
					b--;
					a--;
				}
				if(k>r-n)a--;
				if(b<0||a<0||a<b)break;
				ans=(ans+C(b,a))%MOD;
			}
		}
		printf("%lld\n",ans);
	}
	return 0;
}
