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>
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
using namespace std;
int main(void) {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    long long l, r;
    cin >> l >> r;
    int idx = 0;
    long long sum = 0;
    int lg = -1, rg = -1;
    int ls = -1, rs = -1;
    while ((lg == -1 || rg == -1) && idx < n - 1) {
      long long lsum = sum;
      sum += (long long)(n - 1 - idx) * 2;
      if (lg == -1 && l <= sum) lg = idx, ls = l - lsum;
      if (rg == -1 && r <= sum) rg = idx, rs = r - lsum;
      idx++;
    }
    if (lg == -1) lg = n - 1;
    if (rg == -1) rg = n - 1;
    lg++;
    rg++;
    for (int i = (int)(lg); i <= (int)(rg); i++) {
      if (i == n) {
        cout << "1 ";
        break;
      }
      long long from = 1, to = (long long)(n - i) * 2;
      if (i == lg) from = ls;
      if (i == rg) to = rs;
      for (int j = (int)(from); j <= (int)(to); j++) {
        if (j % 2 == 1) {
          cout << i << " ";
        } else {
          cout << i + j / 2 << " ";
        }
      }
    }
    cout << "\n";
  }
  return 0;
}