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;
template <typename S, typename T>
ostream& operator<<(ostream& out, pair<S, T> const& p) {
  out << '(' << p.first << ", " << p.second << ')';
  return out;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> const& v) {
  long long l = v.size();
  for (long long i = 0; i < l - 1; i++) out << v[i] << ' ';
  if (l > 0) out << v[l - 1];
  return out;
}
template <typename T>
void trace(const char* name, T&& arg1) {
  cout << name << " : " << arg1 << '\n';
}
template <typename T, typename... Args>
void trace(const char* names, T&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  trace(comma + 1, args...);
}
const long long MOD = 1e9 + 7;
const long long INF = 1e18;
const double EPS = 1e-9;
const long long N = 500005;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long rand(long long l, long long r) {
  uniform_int_distribution<long long> uid(l, r);
  return uid(rng);
}
long long n, l, r;
void solve(long long x) {
  cin >> n >> l >> r;
  long long num = 0;
  long long toadd = n - 1;
  long long vert = 1;
  long long to = 2;
  vector<long long> ans;
  for (long long i = 0; i < n - 1; i++) {
    if (2 * (num + toadd) >= l) {
      long long left = 2 * num + 1;
      long long i = 0;
      to = vert + 1;
      while (left <= r and left <= n * (n - 1)) {
        if (i % 2 == 0) {
          if (left >= l) {
            ans.push_back(vert);
          }
        } else {
          if (left >= l) {
            ans.push_back(to);
          }
          to++;
          if (to > n) {
            vert++;
            to = vert + 1;
          }
        }
        left++;
        i++;
      }
      break;
    }
    num += toadd;
    toadd--;
    vert++;
  }
  long long last = (n * (n - 1)) + 1;
  if (r == last) {
    ans.push_back(1);
  }
  for (long long i = 0; i < ans.size(); i++) {
    cout << ans[i] << " ";
  }
  cout << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout.precision(12);
  long long t;
  cin >> t;
  long long i = 0;
  while (i < t) {
    solve(i + 1);
    i++;
  }
}