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 CPP):

Solution:
#include <bits/stdc++.h>
#pragma gcc optimize("O3")
#pragma gcc optimize("Ofast")
using namespace std;
long long n, l, r;
long long i, j, t;
void nx() {
  j++;
  if (j == n + 1) {
    i++;
    j = i + 1;
  }
}
void print() {
  if (!t) {
    cout << i << " ";
  } else {
    cout << j << " ";
    nx();
  }
  t ^= 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int k = 0;
    cin >> n >> l >> r;
    if (r == n * (n - 1) + 1) {
      r--;
      k = 1;
    }
    long long cnt = 1;
    for (i = 1; i <= n && cnt + (n - i) * 2 < l; i++) {
      cnt += (n - i) * 2;
    }
    j = i + 1;
    while (cnt + 2 <= l) {
      cnt += 2;
      j++;
    }
    t = (l - cnt) % 2;
    for (int z = 1; z <= r - l + 1; z++) {
      print();
    }
    if (k)
      cout << 1 << '\n';
    else
      cout << '\n';
  }
}