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 <cstdio>
#include <iostream>
using namespace std;
typedef long long ll;
const ll mod = 1e9 + 7;
inline ll Min (ll x, ll y) {return x < y ? x : y;}
inline ll Max (ll x, ll y) {return x > y ? x : y;}
ll fac[200005];
inline void init (ll n) {
	fac[0] = 1;
	for (ll i = 1; i <= n; i++)
		fac[i] = fac[i - 1] * i % mod;
}
inline ll quickPow (ll a, ll k, ll p) {
	ll res = 1;
	a %= p;
	while (k) {
		if (k & 1)
			res = res * a % p;
		a = a * a % p;
		k >>= 1;
	}
	return res;
}
inline ll inv (ll a, ll p) {
	return quickPow(a, p - 2, p);
}
inline ll getC (ll n, ll m) {
	if (n < m || n < 0 || m < 0)
		return 0;
	return fac[n] * inv(fac[m] * fac[n - m] % mod, mod) % mod;
}
int main () {
	int T;
	scanf("%d", &T);
	init(2e5);
	while (T--) {
		ll n, l, r, ans = 0, tot;
		scanf("%lld %lld %lld", &n, &l, &r);
		if (n & 1)
			tot = (getC(n, n >> 1) + getC(n, (n >> 1) + 1)) % mod;
		else
			tot = getC(n, n >> 1);
		ans = (ans + tot * Min(1 - l, r - n) % mod) % mod;
		for (ll k = Min(1 - l, r - n) + 1;; k++) {
			ll lf = Max(1, l + k);
			ll rg = Min(n, r - k);
			if (rg - lf + 1 < 0)
				break;
			if (n & 1)
				tot = (getC(rg - lf + 1, (n >> 1) - lf + 1) + getC(rg - lf + 1, (n >> 1) + 1 - lf + 1)) % mod;
			else
				tot = getC(rg - lf + 1, (n >> 1) - lf + 1);
			ans = (ans + tot) % mod;
		}
		printf("%lld\n", ans);
	}
}