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 int long long 
const int inf = 1e9;
#define D long double
const long double pi = 3.141592653589793238;
#define FIO ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
const int MOD = 1000000007;
#define fi first
#define se second
#define endl '\n'
const int N = 2e5+10;
int n , l , r ;
int fact[N],ifact[N];

int power(int a, int b){ 
	int x = 1, y = a; 
	while (b > 0){ 
		if (b%2){ 
			x = (x*y)%MOD; 
		} 
		y = (y*y)%MOD; 
		b /= 2; 
	} 
	return x%MOD; 
}

int ncr(int n , int r){
    if(n < r)return 0;
    int ans = fact[n];
    ans %= MOD; ans += MOD; ans %= MOD;
    ans *= ifact[n-r];
    ans %= MOD; ans += MOD; ans %= MOD;
    ans *= ifact[r];
    ans %= MOD; ans += MOD; ans %= MOD;
    return ans;
}

void filler(){
    fact[0] = 1;
    for(int i = 1 ; i < N ; i++){
        fact[i] = i*fact[i-1];
        fact[i] %= MOD; fact[i]+= MOD; fact[i] %= MOD;
    }
    ifact[N-1] = power(fact[N-1], MOD-2);
    for(int i = N-2 ; i >= 0 ; i--){
        ifact[i] = (i+1)*ifact[i+1];
        ifact[i] %= MOD; ifact[i] += MOD; ifact[i] %= MOD;
    } 
    
}


void solve(){
    cin >> n >> l >> r ; 
    int k = min(abs(r-n),abs(l-1));
    if(l-1 >= 0  || r-n <= 0)k = 0;
    // 1--->k all can be -k and +k 
    // from k+1 values will start becoming fixed
    int res = k*ncr(n,n/2);
    // if n is odd then n/2 pos or n/2 neg 
    if(n%2)res += k*ncr(n,n/2);
    res %= MOD ; res += MOD ; res %= MOD;
    //
    k++;
    while(1){
        // fixed -ves
        int fn = 0;
        if(k > r-n)fn = k - (r-n) ;
        // fixed +ves
        int fp = 0;
        if(-k < l-1)fp = l-1 - (-k) ;
        // unfixed 
        int rest = n - fn - fp ; 
        if(rest < 0)break;  
        // neg needed 
        int negn = n/2 - fn ; 
        // pos needed 
        int posn = n/2 - fp ;
        if(negn < -1 || posn < -1)break;
        assert(negn + posn + (n%2) == rest);
        if(n%2 == 0)res += ncr(rest,posn);
        else res += ncr(rest,posn+1) + ncr(rest,posn);
        res %= MOD ; res += MOD ; res %= MOD; 
        k++;
    }
    cout << res << endl;
} 

signed main() {
        FIO
        int t = 1;
        filler();
        cin >> t;
        while(t--){
            solve();
        }

}