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 <cstdio>
#include <iostream>
using namespace std;
const int M = 200005;
const int MOD = 1e9+7;
#define int long long
int read()
{
	int x=0,f=1;char c;
	while((c=getchar())<'0' || c>'9') {if(c=='-') f=-1;}
	while(c>='0' && c<='9') {x=(x<<3)+(x<<1)+(c^48);c=getchar();}
	return x*f;
}
int T,n,m,l,r,ans,fac[M],inv[M];
void init(int n)
{
	fac[0]=inv[0]=inv[1]=1;
	for(int i=2;i<=n;i++) inv[i]=inv[MOD%i]*(MOD-MOD/i)%MOD;
	for(int i=2;i<=n;i++) inv[i]=inv[i-1]*inv[i]%MOD;
	for(int i=1;i<=n;i++) fac[i]=fac[i-1]*i%MOD;
}
int C(int n,int m)
{
	if(n<0 || m<0 || n<m) return 0;
	return fac[n]*inv[m]%MOD*inv[n-m]%MOD;
}
void work()
{
	n=read();l=read();r=read();
	m=n/2;ans=0;
	int t=min(1-l,r-n);
	ans=t*C(n,m)%MOD;
	if(n%2) ans=(ans+t*C(n,m+1))%MOD;
	for(int k=t+1;;k++)
	{
		int lg=max(1ll,k+l),rg=min(n,r-k);
		if(lg>rg+1) break;//attention
		ans=(ans+C(rg-lg+1,m-(lg-1)))%MOD;
		if(n%2) ans=(ans+C(rg-lg+1,m+1-(lg-1)))%MOD;
	}
	printf("%lld\n",(ans+MOD)%MOD);
}
signed main()
{
	T=read();init(2e5);
	while(T--) work();
}