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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  t = 1;
  cin >> t;
  for (int i1 = 0; i1 < t; ++i1) {
    long long n, l, r;
    cin >> n >> l >> r;
    vector<long long> v;
    long long c = (n - 1) * 2;
    long long s = c;
    vector<long long> v1;
    for (int i = 0; i < n - 1; ++i) {
      v.push_back(s);
      v1.push_back(c / 2);
      c -= 2;
      s += c;
    }
    long long x = lower_bound(v.begin(), v.end(), l) - v.begin();
    long long y = lower_bound(v.begin(), v.end(), r) - v.begin();
    if (x == v.size()) {
      cout << 1 << "\n";
      continue;
    }
    long long res1 = v[x] - l + 1;
    if (res1 % 2) {
      res1 /= 2;
      cout << n - res1 << " ";
      l++;
    } else
      res1 /= 2;
    if (x == y) {
      long long dif = r - l + 1;
      while (dif > 1) {
        res1--;
        cout << x + 1 << " " << n - res1 << " ";
        dif -= 2;
      }
      if (dif == 1) cout << x + 1 << "\n";
      continue;
    }
    for (int i = 0; i < res1; ++i) {
      cout << x + 1 << " " << n - res1 + 1 + i << " ";
    }
    x++;
    while (x < y) {
      for (int i = 0; i < v1[x]; ++i) {
        cout << x + 1 << " " << x + i + 2 << " ";
      }
      x++;
    }
    if (y == v.size()) {
      cout << 1 << "\n";
      continue;
    }
    long long res2 = r - v[y - 1];
    res1 = res2;
    res2 /= 2;
    for (int i = 0; i < res2; ++i) {
      cout << y + 1 << " " << y + 2 + i << " ";
    }
    if (res1 % 2) cout << y + 1;
    cout << "\n";
  }
  return 0;
}