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;
void _print(long long int t) { cerr << t; }
void _print(int t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(long double t) { cerr << t; }
void _print(double t) { cerr << t; }
void _print(long long unsigned t) { cerr << t; }
template <class T, class V>
void _print(pair<T, V> p);
template <class T>
void _print(vector<T> v);
template <class T>
void _print(set<T> v);
template <class T, class V>
void _print(map<T, V> v);
template <class T>
void _print(multiset<T> v);
template <class T, class V>
void _print(pair<T, V> p) {
  cerr << "{";
  _print(p.first);
  cerr << ",";
  _print(p.second);
  cerr << "}";
}
template <class T>
void _print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(set<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(multiset<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v) {
  cerr << "[ ";
  for (auto i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class a>
void printarr(a arr[], int n) {
  for (int i = 0; i < n; i++) cout << arr[i] << ' ';
  cout << '\n';
}
template <class c>
void printarr(vector<c> v) {
  for (auto i : v) {
    cout << i << ' ';
  }
  cout << '\n';
}
template <class c>
void printarr(set<c> s) {
  for (auto i : s) {
    cout << i << ' ';
  }
  cout << '\n';
}
template <class t>
t extended_gcd(t a, t b, t &x, t &y) {
  t x1, y1;
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  t g = extended_gcd(b, a % b, x1, y1);
  x = y1;
  y = x1 - (a / b) * y1;
  return g;
}
template <class a>
a powmod(a t1, a t2, a mod) {
  a res = 1;
  t1 = t1 % mod;
  if (t1 == 0) return 0;
  while (t2) {
    if (t2 & 1) res = (res * t1) % mod;
    t1 = (t1 * t1) % mod;
    t2 = t2 >> 1;
  }
  return res;
}
template <class a>
a powmod(a t1, a t2) {
  a res = 1;
  if (t1 == 0) return 0;
  while (t2) {
    if (t2 & 1) res = (res * t1);
    t1 = (t1 * t1);
    t2 = t2 >> 1;
  }
  return res;
}
const long long int MOD = 1e9 + 7;
const int N = 1e2 + 5;
void run() {
  long long int n, l, r;
  cin >> n >> l >> r;
  long long int p = n - 1, last = 0;
  while (p * 2 + last < l) {
    last += p * 2;
    p--;
    if (p == 0) {
      cout << 1 << '\n';
      return;
    }
  }
  long long int t = n - p;
  ;
  long long int tt = t + ceil((l - last) / 2.0);
  tt = (tt == t) ? tt + 1 : tt;
  ;
  last = ((l - last) % 2) ? t : tt;
  while (l <= r) {
    cout << last << ' ';
    if (last == tt) {
      tt++;
      if (tt > n) {
        t++;
        if (t == n) t = 1;
        tt = t + 1;
      }
      last = t;
    } else {
      last = tt;
    }
    l++;
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  cout << setprecision(15) << fixed;
  int test;
  cin >> test;
  while (test--) {
    run();
  }
  cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << '\n';
  return 0;
  ;
}