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]. 
import sys

# sys.setrecursionlimit(200005)
int1 = lambda x: int(x)-1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def LI1(): return list(map(int1, sys.stdin.readline().split()))
def LLI1(rows_number): return [LI1() for _ in range(rows_number)]
def SI(): return sys.stdin.readline().rstrip()
inf = 10**16
md = 10**9+7
# md = 998244353

def nCr(com_n, com_r):
    if com_r < 0: return 0
    if com_n < com_r: return 0
    return fac[com_n]*ifac[com_r]%md*ifac[com_n-com_r]%md

# 準備
n_max = 200005
fac = [1]
for i in range(1, n_max+1): fac.append(fac[-1]*i%md)
ifac = [1]*(n_max+1)
ifac[n_max] = pow(fac[n_max], md-2, md)
for i in range(n_max-1, 1, -1): ifac[i] = ifac[i+1]*(i+1)%md

def solve():
    n, l, r = LI()
    ans = 0
    mn = max(1, 1-r, l-n)
    mx = min(1-l, r-n)
    if mn <= mx:
        if n & 1:
            ans += nCr(n, n//2)*2*(mx-mn+1-(mn <= 0 <= mx))%md
        else:
            ans += nCr(n, n//2)*(mx-mn+1-(mn <= 0 <= mx))%md
    s = mn-1
    # print(mn, mx, ans)
    while s > 0:
        # (-sにできる)[L(どちらにもできる)R](sにできる)
        L = max(1, l+s)
        R = min(n, r-s)
        # print(s, L, R)
        ans += nCr(R-L+1, n//2-L+1)
        if n & 1:
            ans += nCr(R-L+1, n//2-L+2)
        s -= 1
    s = mx+1
    while 1:
        L = max(1, l+s)
        R = min(n, r-s)
        # print(s, L, R)
        if L > R+1: break
        ans += nCr(R-L+1, n//2-L+1)
        if n & 1:
            ans += nCr(R-L+1, n//2-L+2)
        s += 1
        ans %= md
    print(ans)

for testcase in range(II()):
    solve()
