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>
#define M 200010
using namespace std;
int n,l,r,ans,T,h[M],inv[M];
const int mod=1e9+7;
int power(int a,int b) {
	int ans=1;
	while(b) {
		if(b&1) ans=1ll*ans*a%mod;
		b>>=1; a=1ll*a*a%mod;
	}
	return ans;
}
int C(int n,int m) {
	return 1ll*h[n]*inv[m]%mod*inv[n-m]%mod;
}
int In(int x,int l,int r) {
	return (l>=x)+(r>=x);
}
int main() {
	h[0]=inv[0]=1;
	for(int i=1;i<=200000;i++) 
		h[i]=1ll*i*h[i-1]%mod,inv[i]=power(h[i],mod-2);
	scanf("%d",&T);
	while(T--) {
		scanf("%d%d%d",&n,&l,&r);
		int x=min(r-n,1-l);
		ans=1ll*x*C(n,n/2)%mod;
		if(n&1) ans=2ll*ans%mod;
		int h=1,t=n,l1=0,l2=0;
		for(int i=x+1;;i++) {
			int mid=(n+1)/2;
			if(n&1) {
				while(h<=mid&&In(i,r-h,h-l)==1) 
					h++,l1++;
				while(t>=mid&&In(i,r-t,t-l)==1) 
					t--,l2++;
				if(!In(i,r-h,h-l)||!In(i,r-t,t-l))
					break;
			}
			else {
				while(h<=mid&&In(i,r-h,h-l)==1) 
					h++,l1++;
				while(t>mid&&In(i,r-t,t-l)==1) 
					t--,l2++;	
			}
			if(l1>mid||l2>mid) break;
			if(n&1) {
				if(l1==mid||l2==mid) ans=(ans+1)%mod;
				else ans=((ans+C(n-l1-l2,mid-l1))%mod+C(n-l1-l2,mid-l2))%mod;
			}
			else ans=(ans+C(n-l1-l2,mid-l1)%mod)%mod;
			if(l1==mid||l2==mid) break;
		}
		printf("%d\n",ans);
	}
	return 0;
}