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 f0r(a, b) for (long long a = 0; a < b; a++)
#define f1r(a, b, c) for (long long a = b; a < c; a++)
#define f0rd(a, b) for (long long a = b; a >= 0; a--)
#define f1rd(a, b, c) for (long long a = b; a >= c; a--)
#define ms(arr, v) memset(arr, v, sizeof(arr))
#define mp(a, b) make_pair(a, b)
#define pb push_back
#define f first
#define s second

#define ao(a, n) {for (int ele = 0; ele < n; ele++) { if (ele) cout << " "; cout << a[ele]; } cout << '\n';}

typedef long long ll;
typedef double ld;
typedef long double lld;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpi;
typedef vector<pll> vpl;

const ll inf = 1e16;

const int MOD = 1e9 + 7;
typedef decay<decltype(MOD)>::type T;
struct mi {
	T val;
	explicit operator T() const { return val; }
	mi() { val = 0; }
	mi(const long long& v) {
		val = (-MOD <= v && v <= MOD) ? v : v % MOD;
		if (val < 0) val += MOD; }
	friend ostream& operator<<(ostream& os, const mi& a) { return os << a.val; }
	friend bool operator==(const mi& a, const mi& b) { return a.val == b.val; }
	friend bool operator!=(const mi& a, const mi& b) { return !(a == b); }
	friend bool operator<(const mi& a, const mi& b) { return a.val < b.val; }
	mi operator-() const { return mi(-val); }
	mi& operator+=(const mi& m) {
		if ((val += m.val) >= MOD) val -= MOD;
		return *this; }
	mi& operator-=(const mi& m) {
		if ((val -= m.val) < 0) val += MOD;
		return *this; }
	mi& operator*=(const mi& m) { val = (long long) val * m.val % MOD;
		return *this; }
	friend mi pow(mi a, long long p) {
		mi ans = 1; assert(p >= 0);
		for (; p; p /= 2, a *= a) if (p & 1) ans *= a;
		return ans; }
	friend mi inv(const mi& a) { assert(a != 0); return pow(a, MOD - 2); }
	mi& operator/=(const mi& m) { return (*this) *= inv(m); }
	friend mi operator+(mi a, const mi& b) { return a += b; }
	friend mi operator-(mi a, const mi& b) { return a -= b; }
	friend mi operator*(mi a, const mi& b) { return a *= b; }
	friend mi operator/(mi a, const mi& b) { return a /= b; }
};

mi fact[200005];
mi invfact[200005];


mi choose(ll n, ll c) {
	if (c < 0) return 0;
	if (n == 0) return 1;

	return fact[n] * invfact[c] * invfact[n-c];
}


int main() {
	ios_base::sync_with_stdio(false); cin.tie(NULL);

	
	fact[0] = 1;
	invfact[0] = 1;

	for (ll i = 1; i < 200005; i++) {
		fact[i] = fact[i-1] * i;
		invfact[i] = 1/fact[i];
	}

	ll t; cin >> t;
	while (t--) {
		ll n, l, r; cin >> n >> l >> r;

		//find the min k
		ll k = max(1LL, min(1 - l + 1, r - n + 1));

		mi ret = (k-1) * choose(n, n/2);

		if (n % 2) {
			ret *= 2;
		}

		bool pos = 1;

		while (pos) {
			
			ll leftfill = max(0LL, k + l - 1);
			ll rightfill = max(0LL, n - (r - k));

			if (2 * leftfill > n+1 || 2 * rightfill > n+1) {
				pos = 0;
				break;
			}

			ll space = n - leftfill - rightfill;

			if (space == 0) {
				ret += 1;
				break;
			}

			//the guys in the form i + k
			ll type1 = n/2 - leftfill;

			if (n % 2) {
				ret += choose(space, type1) + choose(space, type1 + 1);
			} else {
				ret += choose(space, type1);
			}

			k++;
		}

		cout << ret << endl;



	}



}
