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<algorithm>
#include<vector>
#include<cstdio>

#define N 200100

using namespace std;
typedef long long ll;
const int mod=1e9+7;
//ll two[N]={1,2};
ll inv[N]={1,1};
ll oo[N]={1,1};
ll inva[N]={1,1};
void init()
{
    for(int i=2;i<=200010;i++)
    {
        oo[i]=(oo[i-1]*i)%mod;
        //two[i]=(two[i-1]<<1)%mod;
        inv[i]=inv[mod%i]*(mod-mod/i)%mod;
        inva[i]=inv[i]*inva[i-1]%mod;
    }
}
int cal(int n,int m)
{
    if(m>n||m<0) return 0;
    return (oo[n]*inva[n-m]%mod)*inva[m]%mod;
}
int main()
{
    init();
    int t;scanf("%d",&t);
    int n,l,r;
    while(t--)
    {

        scanf("%d%d%d",&n,&l,&r);
        int k=min(1-l,r-n);
        ll ans=1ll*k*cal(n,n/2)%mod;
        if(n&1) ans=(ans+ans)%mod;
        for(k++;;k++)
        {
            int i=max(0,l+k-1);
            int j=min(n+1,r-k+1);
            if(i>n-n/2||j<=n/2) break;

            ans+=cal(j-i-1,n/2-i);
            if(n&1) ans+=cal(j-i-1,n/2-i+1);
            //ans%=mod;
        }
        ans%=mod;
        printf("%lld\n",ans);
    }
    return 0;
}
