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 sys
def data(): return sys.stdin.buffer.readline().strip()
out=sys.stdout.write
def mdata(): return map(int, data().split())

for t in range(int(data())):
    n,l,r=mdata()
    a=l
    for i in range(1,n+1):
        if 2*(n-i)<=a:
            a-=2*(n-i)
        else:
            break
    cnt=l
    ans=[]
    if l%2==0:
        if a==0:
            ans.append(str(n))
        else:
            ans.append(str(i+a//2))
        cnt+=1
    k=i+a//2+1
    for j in range(i,n):
        while cnt<=r:
            ans.append(str(j))
            cnt+=1
            if cnt>r:
                break
            ans.append(str(k))
            k+=1
            cnt+=1
            if k==n+1:
                k=j+2
                break
    if r==n*(n-1)+1:
        ans.append(str(1))
    out(' '.join(ans)+'\n')