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 T, n;
long long l, r;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> T;
  while (T > 0) {
    --T;
    cin >> n >> l >> r;
    long long s = 0;
    for (int i = 1; i <= n; ++i) {
      long long t = s + 1ll * 2 * (n - i);
      if (t < l) {
        s = t;
        continue;
      }
      if (s < l && t >= l) {
        if (l % 2 == 1) {
          int L = (l - s) / 2 + 1 + i;
          if (t >= r) {
            int R = (r - s) / 2 + i;
            for (int j = L; j <= R; ++j) {
              cout << i << " " << j << " ";
            }
            if ((r - s) % 2 == 1) cout << i << " ";
            s = t;
            break;
          }
          for (int j = L; j <= n; ++j) {
            cout << i << " " << j << " ";
          }
        } else {
          int L = (l - s) / 2 + i;
          cout << L << " ";
          if (t >= r) {
            int R = (r - s) / 2 + i;
            for (int j = L + 1; j <= R; ++j) {
              cout << i << " " << j << " ";
            }
            if ((r - s) % 2 == 1) cout << i << " ";
            s = t;
            break;
          }
          for (int j = L + 1; j <= n; ++j) {
            cout << i << " " << j << " ";
          }
        }
        s = t;
        continue;
      }
      if (t < r) {
        for (int j = i + 1; j <= n; ++j) {
          cout << i << " " << j << " ";
        }
        s = t;
        continue;
      }
      if (s < r && t >= r) {
        int R = (r - s) / 2 + i;
        for (int j = i + 1; j <= R; ++j) {
          cout << i << " " << j << " ";
        }
        if ((r - s) % 2 == 1) cout << i << " ";
        s = t;
      }
    }
    if (s < r) cout << 1;
    cout << '\n';
  }
  return 0;
}