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]. 
def main():
    from sys import stdin, setrecursionlimit
    #from io import BytesIO
    #from os import read, fstat
    #from math import gcd
    #from random import randint, choice, shuffle
    #from itertools import combinations, product
    #from functools import lru_cache
    #from re import search, finditer
    input = stdin.buffer.readline
    #input = BytesIO(read(0, fstat(0).st_size)).readline
    #setrecursionlimit(100000000)
    
    def c(n, k):
        if n < k or k < 0:
            return 0
        return f[n] * rev[k] * rev[n - k] % mod
        
    mod = 10**9 + 7
    f = [1]
    for i in range(1, 200001):
        f.append(f[-1] * i % mod)
    rev = [pow(f[-1], mod - 2, mod)]
    for i in range(200000, 0, -1):
        rev.append(rev[-1] * i % mod)
    rev = rev[::-1]
    for _ in range(int(input())):
        n, l, r = map(int, input().split())
        h = n >> 1
        s = min(1 - l, r - n)
        ans = s * (c(n, h) + [0, c(n, h + 1)][n & 1])
        x = s + 1
        while 1:
            ll = max(1, l + x)
            rr = min(n, r - x)
            if rr - ll + 1 < 0:
                break
            ans += c(rr - ll + 1, h - (ll - 1)) + [0, c(rr - ll + 1, h + 1 - (ll - 1))][n & 1]
            x += 1
        print(ans % mod)
    
main()
