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 <class T, class U>
istream &operator>>(istream &in, pair<T, U> &rhs) {
  in >> rhs.first;
  in >> rhs.second;
  return in;
}
template <class T, class U>
ostream &operator>>(ostream &out, const pair<T, U> &rhs) {
  out << rhs.first;
  out << " ";
  out << rhs.second;
  return out;
}
template <class T, class U>
pair<T, U> operator+(pair<T, U> &a, pair<T, U> &b) {
  return pair<T, U>(a.first + b.first, a.second + b.second);
}
template <class T, class U>
pair<T, U> operator-(pair<T, U> &a, pair<T, U> &b) {
  return pair<T, U>(a.first - b.first, a.second - b.second);
}
template <class T>
istream &operator>>(istream &in, vector<T> &a) {
  for (auto &i : a) cin >> i;
  return in;
}
template <class T>
ostream &operator<<(ostream &out, const vector<T> &a) {
  for (auto &i : a) cout << i << " ";
  return out;
}
template <class T>
ostream &operator<<(ostream &out, const set<T> &a) {
  for (auto &i : a) cout << i << " ";
  return out;
}
template <class T>
ostream &operator<<(ostream &out, const unordered_set<T> &a) {
  for (auto &i : a) cout << i << " ";
  return out;
}
template <class T>
ostream &operator<<(ostream &out, const multiset<T> &a) {
  for (auto &i : a) cout << i << " ";
  return out;
}
template <class T, class U>
ostream &operator<<(ostream &out, const map<T, U> &a) {
  for (auto &i : a) cout << "(" << i.first << ", " << i.second << ")\n";
  return out;
}
template <class T, class U>
ostream &operator<<(ostream &out, const unordered_map<T, U> &a) {
  for (auto &i : a) cout << "(" << i.first << ", " << i.second << ")\n";
  return out;
}
const long long int N = 1e5 + 100;
const long long int inf = 1e9;
const double pi = acos(-1);
void solve(int test_case) {
  long long int n, j = 0, l, r;
  cin >> n >> l >> r;
  long long int cur = 1, done = 0;
  if (l == n * (n - 1) + 1) {
    cout << 1 << "\n";
    return;
  }
  while (done + 2 * (n - cur) < l) {
    done += 2 * (n - cur);
    cur++;
  }
  long long int nxt = cur + 1;
  vector<long long int> ans;
  l -= done;
  while (done <= r) {
    ans.emplace_back(cur);
    done++;
    if (done > r) break;
    ans.emplace_back(nxt);
    nxt++;
    done++;
    if (nxt == n + 1) {
      cur++;
      nxt = cur + 1;
      if (cur == n && done <= r) {
        done++;
        ans.emplace_back(1);
      }
    }
  }
  for (int i = l - 1; i + 1 < ans.size(); ++i) cout << ans[i] << " ";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  int t = 1;
  cin >> t;
  for (int i = 1; i < t + 1; i++) solve(i);
}