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 math
import sys
from collections import defaultdict,Counter,deque,OrderedDict
import bisect
#sys.setrecursionlimit(1000000)
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
ilele = lambda: map(int,input().split())
alele = lambda: list(map(int, input().split()))
def list2d(a, b, c): return [[c] * b for i in range(a)]
#def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
#INF = 10 ** 18
#MOD = 1000000000 + 7
from itertools import accumulate,groupby

for _ in range(int(input())):
    n,l,r = ilele()
    z = n*(n-1)+ 1
    if (l==r and l == z) or (r==1) :
        print(1)
        continue
    k = n-1;tot = 0;f=0
    while tot < l and k>0:
        f+=1
        tot += 2*k
        k-=1
        
    #print(f,tot)
    tot -= 2*(k+1)
    tot += 1
    Ans = []
    g=0;C=[];m=0;h = f+1;i=tot
    while i<r+1 and i<z :
        if g==0:
            if i==l:
                g=1
            if m%2==0:
                C.append(f)
            else:
                if h<n:
                    C.append(h)
                    h+=1
                else:
                    C.append(h)
                    f+=1
                    h = f+1
            if g==1:
                Ans.append(C[-1])
        elif g==1:
            if m%2==0:
                Ans.append(f)
            else:
                if h<n:
                    Ans.append(h)
                    h+=1
                else:
                    Ans.append(h)
                    f+=1
                    h = f+1
        m+=1
        i+=1
    if r==z:
        Ans.append(1)
    #print(C,Ans)
    print(*Ans)