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;
#define ll long long

const ll mod=1e9+7;
const ll maxn = 2e5+5;

ll fact[maxn+1];
ll invfact[maxn+1];

ll modexp(ll a,ll b){
    ll ans = 1;
	while (b > 0) {
		if (b & 1)
			ans = (a*ans)%mod;
		a = (a*a)%mod;
		b = b>>1;
	}
	return ans;
}

void preprocess(){
	ll p=mod;
	fact[0]=1;
	ll i;
	for(i=1;i<maxn;i++){
		fact[i]=(i*fact[i-1])%p;
	}
	i--;
	invfact[i]=modexp(fact[i],p-2);
	for(i--;i>=0;i--){
		invfact[i]=(invfact[i+1]*(i+1))%p;
	}
}

ll ncr(ll x,ll y){

    if(x<y || x<0 || y<0){
        return 0;
    }

    return ((fact[x]*invfact[y])%mod * invfact[x-y])%mod;

}

int main(){

    preprocess();

    ll nt;
    cin>>nt;

    while(nt--){

        ll n,l,r;
        cin>>n>>l>>r;

        ll ans = 0;
        ll allpos = min( - (l-1),r-n);
        ans=(allpos*ncr(n,n/2))%mod;
        if(n & 1){
            ans=(ans + allpos*ncr(n,n-n/2))%mod;
        }
        //cout<<ans<<endl;
        for(ll i=0;i<n;i++){

            ll c_val = allpos + i + 1;
            ll p_poss = min(n,r-c_val);
            ll n_poss = min(n,n - (l+c_val) + 1);
            if(p_poss<0 || n_poss<0){
                break;
            }
            ll both_poss = p_poss + n_poss - n;
            ll only_pos = p_poss - both_poss;
            ll only_neg = n_poss - both_poss;

            ans = (ans + ncr(both_poss,n/2-only_pos)%mod)%mod;
            if(n & 1){
                ans = (ans + ncr(both_poss,n/2 - only_neg)%mod)%mod;
            }

            //cout<<c_val<<" "<<i<<" "<<ans<<endl;

        }

        cout<<ans<<"\n";

    }


    return 0;
}
