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:
#code
import sys
import math as mt
#input=sys.stdin.buffer.readline
t=int(input())
#tot=0
for __ in range(t):
    #n=int(input())      
    #l=list(map(int,input().split()))  
    n,l,r=map(int,input().split())            
    j=1
    k=2*n-2
    mul=1
    k=2*n-2
    r1=k
    l1=1
    for i in range(n-2):
        if l>=l1 and l<=r1:
            #print(111,l1,r1,mul)
            break
        k-=2
        l1=r1+1
        r1=l1+k-1
        mul+=1
        
    #print(111,l1,r1,mul,k)
    nex=mul
    ch=0
    #print(999,ch)
    for i in range(l1,min(r1+1,r+1)):
        if i>=l:
            if ch%2!=0:
                nex+=1
                print(nex,end=" ")
            else:
                print(mul,end=" ")
        else:
            if ch%2!=0:
                nex+=1
        ch+=1
        if i>r1:
            break
    i=r1+1
    mul+=1
    ch=0
    nex=mul
    k-=1
    while i<=min(r,n*(n-1)):
        if ch%2==0:
            print(mul,end=" ")
        else:
            nex+=1
            print(nex,end=" ")
        i+=1
        ch+=1
        if ch==2*(n-mul):
            mul+=1
            ch=0
            nex=mul
            
    if r==n*(n-1)+1:
        print(1,end=" ")
    print()