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]. 
mod = 1000000007
f = [1] * 200001
for i in xrange(2, 200001):
    f[i] = f[i-1] * i % mod
invf = [1] * 200001
invf[-1] = pow(f[-1], mod - 2, mod)
for i in xrange(200000, 2, -1):
    invf[i-1] = invf[i] * i % mod
def C(n, k):
    if k < 0 or k > n:
        return 0
    return f[n] * invf[k] * invf[n-k] % mod
def solve(mn=min, mx=max):
    n, l, r = map(int, raw_input().split())
    p = min(r - n, 1 - l)
    ans = C(n, n / 2) * p % mod
    if n % 2:
        ans = ans * 2 % mod
    while 1:
        p += 1
        u = n - mn(n, r - p)
        d = mx(l + p, 1) - 1
        if u + d > n:
            break
        ans += C(n - u - d, n / 2 - u)
        if n % 2:
            ans += C(n - u - d, n - n / 2 - u)
        ans %= mod
    print ans

T = int(raw_input())
for t in xrange(T):
    solve()
