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:
t = int(input())
for i in range(t):
    n, l, r = map(int, input().split())
    if l == n * (n - 1) + 1:
        print(1)
    else:
        x = 1
        summa = x * 2 * n
        rasn = x * (x + 1)
        while summa - rasn < l:
            summa += 2 * n
            rasn = (rasn // x) * (x + 2)
            x += 1
        x -= 1
        summa -= 2 * n
        rasn = (rasn // (x + 2)) * x
        first = x + 1
        second = (l - summa + rasn + 1) // 2 + first
        if (l - summa + rasn) % 2 == 0:
            print(second, end=" ")
            if second == n:
                first += 1
                second = first + 1
            else:
                second += 1
            ind = l + 1
        else:
            ind = l
        while ind + 1 <= r:
            print(first, second, end=" ")
            if second == n:
                first += 1
                second = first + 1
            else:
                second += 1
            ind += 2
        if ind == r:
            if r == n * (n - 1) + 1:
                print(1)
            elif second == n:
                print(first)
            else:
                print(first)
        else:
            print()