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

def allWays(start, verts, done, stack, n):
    global valid
    if not valid:
        return
    stack.append(start)
    if len(done) == len(verts):
        print(stack)
        stack.pop()
        valid = False
        return
    for x in range(1,n+1):
        if start != x and not (start,x) in done:
            done.add((start,x))
            allWays(x,verts,done,stack,n)
            done.remove((start,x))
    stack.pop()
'''
for y in range(1,10):
    verts = set([((x//y) + 1, (x%y)+1) for x in range(y**2)])
    for x in range(1,y+1):
        verts.remove((x,x))
    valid = True
    print(y, end=' ')
    allWays(1, verts, set(), [], y)
'''

def order(n,x):
    out = []
    for y in range(x+1,n+1):
        out.append(x)
        out.append(y)
    return out

for case in range(int(stdin.readline())):
    n,l,r = [int(x) for x in stdin.readline().split()]

    end1 = False

    if r == n*(n-1) + 1:
        end1 = True
        r -= 1

    if l == n*(n-1) + 1:
        print(1)
    

    else:
        x = 1

        while l > 2*(n-x):
            l -= 2*(n-x)
            r -= 2*(n-x)
            x += 1

        out = order(n,x)
        r -= 2*(n-x)
        x += 1

        while r > 0:
            out += order(n,x)
            r -= 2*(n-x)
            x += 1

        if end1:
            out += [1]

        if r != 0:
            realOut = out[l-1:r]
        else:
            realOut = out[l-1:]
        print(' '.join([str(b) for b in realOut]))