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 
t=int(sys.stdin.readline())
for _ in range(t):
    n,l,r=map(int,sys.stdin.readline().split())
    prev=0
    cur=0
    start=1
    if l==r and l==n*(n-1)+1:
        print(1)
    else:
        ans=[]
        while(True):    
            cur+=(n-start)*2
            if l<=cur:
                pos=l-prev
                total=r-l+1
                if(pos%2==1):
                    ans.append(start)
                    total-=1 
                    x=start+pos//2 +1
                    while(total>0):
                        ans.append(x)
                        if x==n:
                            start+=1 
                            if start==n:
                                start=1 
                            x=start
                        total-=1 
                        if total>0:
                            ans.append(start)
                            total-=1
                            x+=1 
                else:
                    x=start+pos//2
                    while(total>0):
                        ans.append(x)
                        if x==n:
                            start+=1 
                            if start==n:
                                start=1 
                            x=start
                        total-=1 
                        if total>0:
                            ans.append(start)
                            total-=1
                            x+=1
                break
            prev=cur
            start+=1
        print(*ans)