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 int N = 100005;
void solve() {
  long long int i, j, k, n, m, ans = 0, cnt = 0, sum = 0;
  long long int l, r;
  cin >> n >> l >> r;
  vector<long long int> temp;
  for (i = 1; i < n; i++) {
    temp.push_back(2 * (n - i));
  }
  temp.push_back(1);
  m = temp.size();
  cnt = 1;
  for (i = 0; i < m; i++) {
    if (cnt + temp[i] > l) {
      break;
    } else {
      cnt += temp[i];
    }
  }
  i++;
  if (i > n) {
    i = 1;
  }
  j = i + 1;
  ans = 0;
  long long int cur = i;
  while (cnt <= r) {
    if (cnt == n * (n - 1) + 1) {
      cout << 1;
      break;
    }
    if (cnt >= l) {
      cout << cur << " ";
    }
    if (ans == 0) {
      cur = j;
      j++;
    } else {
      if (j == n + 1) {
        i++;
        j = i + 1;
      }
      cur = i;
    }
    ans ^= 1;
    cnt++;
  }
  cout << '\n';
  return;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) solve();
  return 0;
}