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<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=200010;
const int mod=1e9+7;
int T,n,l,r,a[N],h,t,x,y;
ll ans,fac[N],inv[N],now,pre;
ll qpow(ll x,ll y)
{
	ll res=1;
	while(y)
	  {
	  	if(y&1)res=res*x%mod;
	  	x=x*x%mod;
	  	y>>=1;
	  }
	return res;
}
ll C(ll n,ll m)
{
	if(n<0||m<0)return 0;
	if(m>n)return 0;
	return fac[n]*inv[n-m]%mod*inv[m]%mod;
}
int main()
{
	fac[1]=1;
	for(int i=2;i<=200000;++i)fac[i]=fac[i-1]*i%mod;
	inv[200000]=qpow(fac[200000],mod-2);
	for(int i=199999;i;--i)inv[i]=inv[i+1]*(i+1)%mod;
	inv[0]=1;
	scanf("%d",&T);
	while(T--)
	  {
	  	scanf("%d%d%d",&n,&l,&r);
	  	x=n>>1;if(n&1)y=x+1;else y=x;
	  	h=1;t=n;ans=now=pre=0;
		while(h<=t)
		  {
		  	now=min(h-l,r-t);
		  	(ans+=(now-pre)*C(t-h+1,x-h+1)%mod)%=mod;
		  	if(x!=y)(ans+=(now-pre)*C(t-h+1,y-h+1)%mod)%=mod;
		  	if(h-l<r-t)++h;
		  	else if(h-l>r-t)--t;
		  	else ++h,--t;
		  	pre=now;
		  }
		//printf("%d %d %d %d %d\n",ans,now,pre,h,t);
		if(h-t==1&&(t==x||t==y))ans+=min(r-t,h-l)-pre;
		printf("%d\n",ans);
	  }
}