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:
from sys import stdin, gettrace
from math import sqrt

if not gettrace():
    def input():
        return next(stdin)[:-1]


# def input():
#    return stdin.buffer.readline()

def main():
    def solve():

        n,l,r = map(int, input().split())
        lv = int((2*n+1 - sqrt((2*n-1)**2 -4*(l-1)))/2)
        lvs = -2*n+2*n*lv-lv*lv+lv
        lrd = l - lvs - 1
        res = []
        i = lv
        j = lv+lrd//2 + 1
        if l%2 == 0:
            res = [j]
            if j < n:
                j+=1
            else:
                i+=1
                j = i+1
        for _ in range(l-1, r, 2):
            res += [i,j]
            if j < n:
                j += 1
            else:
                i +=1
                j = i+1
        if r == n*(n-1)+1:
            res[r-l] = 1
        print(*res[:r-l+1])



    q = int(input())
    for _ in range(q):
        solve()


if __name__ == "__main__":
    main()