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 PYTHON):

Solution:
from sys import stdout
from bisect import bisect_left as bl

# large output
out = []

def solve(l, r, n):
    if l == n * (n - 1) + 1:
        out.append('1')
        return
    
    occs = [i for i in xrange(n - 1, 0, -1)]
    
    pref = [occs[i] for i in xrange(n - 1)]
    for i in xrange(1, n - 1): pref[i] = pref[i - 1] + occs[i]

    # pair number
    k = (l + 1) // 2
    
    # find the kth pair
    idx = bl(pref, k)

    # first element of each pair
    a = n - occs[idx]

    relpos = k
    if idx > 0: relpos -= pref[idx - 1]

    # second element of this pair
    b = a + relpos

    # time to print the answer
    ans = []
    cnt = r - l + 1

    x, y = a, b
    
    if l % 2 == 0:
        ans.append(str(b))
        cnt -= 1

        x, y = a, b + 1
    
    if y > n:
        x += 1
        y = x + 1

    if x == n: x = 1
        
    while cnt > 0:
        if cnt >= 2:
            ans.append(str(x))
            ans.append(str(y))
            cnt -= 2
        elif cnt == 1:
            ans.append(str(x))
            cnt -= 1

        y += 1
        if y > n:
            x += 1
            y = x + 1

        if x == n: x = 1

    out.append(' '.join(ans))

def main():
    for _ in xrange(input()):
        n, l, r = map(int, raw_input().strip().split())
        solve(l, r, n)

    stdout.write('\n'.join(out))

if __name__ == "__main__": main()