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>
#define del(h) while(h.size() && h.top().second <l) h.pop()
using namespace std;
//FILE *fi = freopen("1.txt","r",stdin);
const long long Mod = 1e9 + 7;
typedef pair <int,int> ii;
const int Maxn = 2e5 + 10;
long long g[Maxn];
long long ig[Maxn];
long long power(long long n,int x)
{
    if(!x)
        return 1;
    if(x & 1)
        return n * power(n * n % Mod,x/2) % Mod;
    return power(n*n%Mod,x/2);
}
long long Nck(int n,int k)
{
    if(k > n || n < 0 || k < 0)
        return 0;
    return g[n] * ig[k] % Mod * ig[n-k] % Mod;
}
int main()
{
    iostream::sync_with_stdio(false);
    cin.tie(0);
    g[0] = 1;
    for(int i = 1;i < Maxn;i++)
        g[i] = g[i-1] * i % Mod;
    ig[Maxn-1] = power(g[Maxn-1],Mod-2);
    for(int i = Maxn-2;i>=0;i--)
    {
        ig[i] = ig[i+1] * (i + 1) % Mod;
    }
    int t;
    cin>>t;
    while(t--)
    {
        int n,l,r;
        cin>>n>>l>>r;
        long long ans = 0;
        long long k = min(1 - l,r - n);
        ans += Nck(n,n/2) % Mod;
        if(n & 1)
            ans += Nck(n,n/2) % Mod;
        ans *= k;
        ans %= Mod;
        int K = k;
        int R = n;
        int L = 1;
        for(int X = k + 1;X<=n + k;X++)
        {
            while(R >= 1 && X > r - R)
                R--;
            while(L <= n && -X < l - L)
                L++;
            int x = L - 1;
            int y = n - R;
            (ans += Nck(n - (x + y),n/2 - x)) %= Mod;
            if(n & 1)
                (ans += Nck(n - (x + y),n/2 - y)) %= Mod;

        }
        cout<<ans<<"\n";
    }
    return 0;
}
