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 N 200007
#define mod 1000000007ll
using namespace std;
typedef long long ll;

inline int rd() {
	int x = 0;
	bool f = 0;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) f |= (c == '-');
	for (; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
	return f ? -x : x;
}

int fac[N], ifac[N];

inline int fpow(int x, int t) {
	int res = 1;
	for (; t; t >>= 1, x = 1ll * x * x % mod)
		if (t & 1) res = 1ll * res * x % mod;
	return res;
}

inline void init() {
	fac[0] = ifac[0] = 1;
	for (int i = 1; i < N; ++i) 
		fac[i] = 1ll * fac[i - 1] * i % mod;
	ifac[N - 1] = fpow(fac[N - 1], mod - 2);
	for (int i = N - 2; i; --i) 
		ifac[i] = 1ll * ifac[i + 1] * (i + 1) % mod;
}

inline int C(int n, int m) {
	return 1ll * fac[n] * ifac[m] % mod * ifac[n - m] % mod;
}

inline void work() {
	int n = rd();
	int l = rd();
	int r = rd();
	ll ans = 1ll * C(n, n / 2) * min(1 - l, r - n) % mod;
	int lim = min(1 - l, r - n);
	if (n & 1) ans = ans * 2 % mod;
	for (int t = lim + 1, L = 1, R = n; ; ++t) {
		while(L - t < l) ++L;
		while(R + t > r) --R;
		if (n - L + 1 < n / 2 || R < n - n / 2) break;
		ans = (ans + C(R - L + 1, n / 2 - (n - R))) % mod;
	}
	if (n & 1) {
		for (int t = lim + 1, L = 1, R = n; ; ++t) {
			while(L - t < l) ++L;
			while(R + t > r) --R;
			if (n - L + 1 < n / 2 + 1 || R < n / 2) break;
			ans = (ans + C(R - L + 1, n / 2 + 1 - (n - R))) % mod;
		}
	}
	printf("%lld\n", ans);
}

int main() {
	init();
	int t = rd();
	while (t--) work();
	return 0;
}