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)
        continue
    left, right, summ = 0, n, 0
    while (left != right - 1):
        mid = (left + right) // 2;
        tmp = n * mid - (mid * (mid + 1) // 2)
        if (2 * tmp < l):
            left = mid
            summ = 2 * tmp
        else:
            right = mid
        #print(left, right)
    beg = summ
    count = left + 1
    ans = []
    ind = False
    for i in range(count, n):
        for j in range(i + 1, n + 1):
            ans.append(i)
            ans.append(j)
            summ += 2
            if (summ > r):
                ind = True
                break
        if (ind):
            break
    ans.append(1);
    for i in range(l - 1, r):
        print(ans[i - beg], end = " ")
    print()