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:
def getK(index, N):
    k = 0
    tot = 0
    while True:
        k = k + 1
        alt = tot + 2 * (N -k)
        #print(alt, index)
        if alt < index:
            tot = alt
        else:
            return k, tot
    

def getLoop(k, indexInLoop):
    if indexInLoop % 2 == 0:
        return k
    return indexInLoop//2 + k + 1


T = int(raw_input())
for t in range(T):
    N, L, R = [int(v) for v in raw_input().split()]
    if L == N * (N-1) +1:
        print('1')
    else:
        k, tot = getK(L, N)
        indexInLoop = L - tot- 1
        i = 0
        out = []
        R2 = R
        if R == N * (N-1) + 1:
            R2 = R -1

        while i < R2 - L + 1:
            res = getLoop(k, indexInLoop)
            indexInLoop += 1
            if indexInLoop >= 2 * (N -k):
                k += 1
                indexInLoop = 0
            out.append(str(res))

            i += 1
        if R == N * (N-1) + 1:
            out.append('1')

        print(' '.join(out))