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 check(x,n,l):
    val = 2*n*x-(x*x)-x
    if val < l:
        return True

    return False

def solve(n,l,r,ans):
    low = 1
    high = n-1
    x = 0
    while low <= high:
        mid = (low+high)//2
        if check(mid,n,l):
            x = mid
            low = mid+1
        else:
            high = mid-1

    val = 2*n*x-(x*x)-x
    rem = l-val

    #print(x)
    temp = True
    prev = True
    for i in range(rem-1):
        if prev:
            if temp:
                x += 1
                y = x
        else:
            y += 1
            temp = False
            
        if y == n:
            temp = True

        prev = not prev

        #print(x,y)

    #print(temp,prev,y)
    arr = []
    for i in range(r-l+1):
        if prev:
            if temp:
                x += 1
                y = x

            if x == n:
                x = 1
            arr.append(x)
        else:
            temp = False
            y += 1
            arr.append(y)
            
        if y == n:
            temp = True

        prev = not prev

    ans.append(arr)

def main():
    t = int(input())
    ans = []
    for i in range(t):
        n,l,r = map(int,input().split())
        solve(n,l,r,ans)

    for i in ans:
        for j in i:
            print(j,end = ' ')

        print()


main()