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:
t=int(input())
for  _ in range(t):
    n,l,r=map(int,input().split())
    L=[0]
    tt=2*(n-1)
    for i in range(n):
        L.append(tt)
        tt-=2
    L[-1]=1

    temp=0
    ct=r-l+1
    c=0

    tot=0
    for i in range(1,len(L)):
        if(tot+L[i]<l):
            tot+=L[i]
        else:
            rem=l-tot
            p1=i
            p2=rem//2
            if(rem%2!=0):
                p2+=1
                temp=1
            p2+=i
            break

    if(temp==0):
        print(p2,end=" ")
        p2+=1
        if(p2==n+1):
            p1+=1
            if(p1==n):
                p1=1
            p2=p1+1
        c+=1
    if(p1==n):
        p1=1
    while(c<ct):
        print(p1,end=" ")
        c+=1
        if(c==ct):
            break
        print(p2,end=" ")
        c+=1
        p2+=1
        if(p2==n+1):
            p1+=1
            if(p1==n):
                p1=1
            p2=p1+1


    print(" ")