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:
def calcStartIndex(vertex,n):
    i=vertex
    return 1+2*(i-1)*n-i*i+i

def main():
    
    t=int(input())
    allans=[]
    for _ in range(t):
        n,l,r=readIntArr()
        
        startVertex=1
        b=n
        while b>0:
            while startVertex+b<=n and calcStartIndex(startVertex+b,n)<=l:
                startVertex+=b
            b//=2
        
        sv=startVertex
        idx=calcStartIndex(sv,n)
        ans=[]
        adder=1
        addTurn=False
        while idx<=r:
            #sv,sv+1,sv,sv+2,...sv,n. then sv+=1. if sv==n, then put 1 instead of sv
            if addTurn:
                curr=sv+adder
                adder+=1
            else:
                curr=sv
            
            if idx>=l:
                if sv<n:
                    ans.append(curr)
                else:
                    ans.append(1)
            
            addTurn=not addTurn
            idx+=1
            if curr==n:
                sv+=1
                adder=1
                addTurn=False
        
        allans.append(ans)
    multiLineArrayOfArraysPrint(allans)
            
    
    return
    
#import sys
#input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
import sys
input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
 
def oneLineArrayPrint(arr):
    print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
    print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
    print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
 
def readIntArr():
    return [int(x) for x in input().split()]
 
inf=float('inf')
MOD=10**9+7
 
main()