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:
import math
import sys

A = 1
B = 1
C = 1 
K = 0

def LFromI(I):
    return I*I*A + I*B + C

def IFromL(L):
    return math.floor((-B+math.sqrt(B*B - 4*A*(C-L)))/(2*A))

def Out(l, R):
    curint = IFromL(l)
    curpos = LFromI(curint)

    while curpos <= R:
        if curint == K:
            print(1, end=' ')
            break
        for i in range(2*(K-curint)):
            if curpos+i > R:
                break
            if curpos+i < l:
                continue
            if i % 2 == 0:
                print(curint, end=' ')
            else:
                print(curint + (i+1)//2, end=' ' )
        curint += 1
        curpos = LFromI(curint)
    return 1;

t = int(input())

for _ in range(t):
    K, l, r = map(int, input().split())

    A = -1
    B = 2*K+1
    C = -2*K

    Out(l-1, r-1)
    print()