Problem: You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.

You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).

We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 — a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.

Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.

Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.

Input

The first line contains the single integer T (1 ≤ T ≤ 100) — the number of test cases.

Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — the number of vertices in K_n, and segment of the cycle to print.

It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.

Output

For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.

Example

Input


3
2 1 3
3 3 6
99995 9998900031 9998900031


Output


1 2 1 
1 3 2 3 
1 

Note

In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.

In the third test case, it's quite obvious that the cycle should start and end in vertex 1.

Solution (in PYTHON3):

Solution:
# problem 2
def calc(N, index):
    # find the number when given (N,index)
    if index == 1 or index == N*(N-1)+1:
        return 1
    else:
        wave_l = 1
        wave_r = N
        while wave_r-wave_l > 1:
            M = (wave_l+wave_r)//2
            if (2*N-M)*(M-1) < index:
                wave_l = M
            else:
                wave_r = M
        wave = wave_l
        Left = index-(2*N-wave)*(wave-1)
        if Left % 2 != 0:
            return wave
        else:
            return wave+Left//2

    # maximum value := cnt
    Left = index-(cnt-1)*(cnt-2)
    if Left % 2 == 0:
        return cnt
    else:
        if cnt != Left//2+1:
            return Left//2+1
        else:
            return 1


def solve():
    n, l, r = map(int, input().split())
    A = []
    for i in range(l, r+1):
        A.append(calc(n, i))
    print(*A)


def main():
    T = int(input())
    for i in range(T):
        solve()


if __name__ == "__main__":
    main()