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;
const int mod = 1e9+7;
const int N = 200005;
int n,l,r,ans,pw[N];
int inv(int x) {
	int y = mod-2,sum = 1;
	while(y) {
		if(y&1) sum = 1ll*sum*x%mod;
		y>>=1;
		x=1ll*x*x%mod;
	}
	return sum;
}
int C(int n,int m) {
	return 1ll*pw[n]*inv(pw[m])%mod*inv(pw[n-m])%mod;
}
int main() {
	int T;scanf("%d",&T);
	pw[0]=1;
	for(int i=1;i<N;i++) pw[i] = 1ll*pw[i-1]*i%mod;
	while(T--) {
		scanf("%d%d%d",&n,&l,&r);
		if(r<n||l>1||r-l+1<n) {puts("0");continue;}
		int L = max(l-1,n-r), R = min(r-n,1-l);
		ans = 1ll*R*C(n,n/2)%mod;
		if(n&1) ans=1ll*ans*2%mod;
		int ori1 = 0, ori2 = 0;
		for(int i=R+1;ori1<=n/2&&ori2<=n/2;i++) {
			if(i>r-n) ori1++;
			if(i>1-l) ori2++;
			if(ori1+ori2>n) break;
			if(ori1==(n+1)/2||ori2==(n+1)/2) {ans=(ans+1)%mod;break;}
			if(n&1) ans=(1ll*ans+C(n-ori1-ori2,n/2-ori1)+C(n-ori1-ori2,n/2+1-ori1))%mod;
			else ans = (1ll*ans+C(n-ori1-ori2,n/2-ori1))%mod;
		}
		printf("%d\n",ans);
	}
	return 0;
}