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 io
import os

from collections import Counter, defaultdict, deque


def solve(N, L, R):
    M = R - L + 1
    ans = []
    count = 0
    done = False
    for i in range(N - 1, -1, -1):
        if count + 2 * i < L:
            count += 2 * i
        else:
            curr = N - i
            for j in range(curr + 1, N + 1):
                if count + 1 >= L:
                    ans.append(str(curr))
                    if len(ans) == M:
                        done = True
                        break
                count += 1
                if count + 1 >= L:
                    ans.append(str(j))
                    if len(ans) == M:
                        done = True
                        break
                count += 1
            if done:
                break

    if len(ans) < M:
        ans.append("1")

    return " ".join(ans)


if __name__ == "__main__":
    input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

    T = int(input())
    for t in range(T):
        N, L, R = [int(x) for x in input().split()]
        ans = solve(N, L, R)
        print(ans)