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 sys;input=sys.stdin.readline
t, = map(int, input().split())

def gen(k):
    return [k+(i+1)//2 if i%2 else k for i in range(2*(n-k))]

for _ in range(t):
    n, s, e = map(int, input().split())
    k, sm = 0, 0
    while sm < s and k < n-1:
        k += 1
        sm += 2*(n-k)
    if sm < s:
        k+=1

    k2, sm2 = 0, 0
    while sm2 < e and k2 < n-1:
        k2 += 1
        sm2 += 2*(n-k2)
    if sm2 < e:
        k2+=1
    if k == n:
        print(1)
    elif k == k2:
        L = gen(k)
        tt, tt2 = sm-2*(n-k), sm2-2*(n-k2)
        print(*L[s-tt-1:e-tt2])
    else:
        tt, tt2 = sm-2*(n-k), sm2-2*(n-k2)
        L = gen(k)[s-tt-1:]
        L2 = gen(k2)[:e-tt2]
        for l in range(k+1, k2):
            L += gen(l)
        L += L2
        if k2 == n:
            L.append(1)
        print(*L)