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
readline = sys.stdin.readline

def overlap(a, b, c, d):
    if b <= c or d <= a:
        return False
    return True

T = int(readline())
Ans = [None]*T

for qu in range(T):
    N, l, r = map(int, readline().split())
    l -= 1
    ans = []
    num =[0] + [2*(N-i-1) for i in range(N-1)] + [1]
    for i in range(1, N+1):
        num[i] += num[i-1]
    started = False
    ended = False
    for j in range(N):
        xp = num[j]
        xn = num[j+1]
        if overlap(xp, xn, l, r):
            started = True
        if started:
            cnt = j+2
            for k in range(2*(N-j-1)):
                if xp + k + 1 <= l:
                    if k & 1:
                        cnt += 1
                    continue
                if xp + k + 1 <= r:
                    if k % 2 == 0:
                        ans.append(j+1)
                    else:
                        ans.append(cnt)
                        cnt += 1
                else:
                    ended = True
                    break
        if ended:
            break
                
    if r == N*(N-1)+1:
        ans.append(1)
    Ans[qu] = ' '.join(map(str, ans))
print('\n'.join(map(str, Ans)))