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>
using namespace std;
const long long INF = (long long)2e9;
const long long inf = (long long)2e18;
const long double eps = (long double)1e-8;
const long long mod = (long long)1e9 + 7;
const long long MAXN = (long long)1e5 + 1;
const long long MAXC = (long long)1e6 + 1;
const long long MAXE = (long long)1000;
const long long MAXLOG = 21;
const long long maxlen = (long long)1e5;
const long long asci = (long long)256;
const long long block = 480;
const long double PI = acos(-1);
const long double e = 2.7182818284;
template <class T>
istream &operator>>(istream &in, vector<T> &arr) {
  for (T &cnt : arr) {
    in >> cnt;
  }
  return in;
};
void solve() {
  long long n, l, r;
  cin >> n >> l >> r;
  l--, r--;
  vector<long long> a;
  long long i = 0;
  for (; i < n - 1; ++i) {
    if (l >= 2 * (n - i - 1))
      l -= 2 * (n - i - 1), r -= 2 * (n - i - 1);
    else
      break;
  }
  while (a.size() <= r && i < n) {
    for (long long j = i + 1; j < n; ++j) {
      a.push_back(i);
      a.push_back(j);
    }
    ++i;
  }
  if (i == n) a.push_back(0);
  for (; l <= r; ++l) {
    cout << a[l] + 1 << " ";
  }
  cout << "\n";
}
int main() {
  srand(time(0));
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout.precision(30);
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}