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::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, lgrp = 0, rgrp = 0;
    long long int l, r;
    cin >> n >> l >> r;
    for (int k = 1; k <= n - 1; ++k) {
      if (l <= 2 * (n - k) * 1LL) {
        lgrp = k;
        break;
      } else {
        l -= 2 * (n - k) * 1LL;
      }
    }
    for (int k = 1; k <= n - 1; ++k) {
      if (r <= 2 * (n - k) * 1LL) {
        rgrp = k;
        break;
      } else {
        r -= 2 * (n - k) * 1LL;
      }
    }
    int start = 0;
    if (lgrp == 0) lgrp = n;
    if (rgrp == 0) rgrp = n;
    if (lgrp == n && rgrp == n) {
      cout << 1;
    } else if (lgrp != rgrp) {
      if (l % 2 == 1) {
        ++l;
        cout << lgrp << " ";
      }
      start = l / 2;
      cout << lgrp + start << " ";
      ++l;
      for (int i = l; i <= 2 * (n - lgrp); ++i) {
        if (i % 2 == 0) {
          ++start;
          cout << lgrp + start << " ";
        } else {
          cout << lgrp << " ";
        }
      }
      for (int i = lgrp + 1; i < rgrp; ++i) {
        start = 0;
        for (int j = 1; j <= 2 * (n - i); ++j) {
          if (j % 2 == 0) {
            ++start;
            cout << i + start << " ";
          } else {
            cout << i << " ";
          }
        }
      }
      if (rgrp == n) {
        cout << 1 << " ";
      } else {
        start = 0;
        for (int i = 1; i <= r; ++i) {
          if (i % 2 == 0) {
            ++start;
            cout << rgrp + start << " ";
          } else {
            cout << rgrp << " ";
          }
        }
      }
    } else {
      if (l == r) {
        if (l % 2 == 1) {
          cout << lgrp << " ";
          ++l;
        } else {
          start = l / 2;
          cout << lgrp + start << " ";
          ++l;
        }
      } else if (l % 2 == 1) {
        ++l;
        cout << lgrp << " ";
      }
      if (l <= r) {
        start = l / 2;
        cout << lgrp + start << " ";
        ++l;
      }
      for (int i = l; i <= r; ++i) {
        if (i % 2 == 0) {
          ++start;
          cout << lgrp + start << " ";
        } else {
          cout << lgrp << " ";
        }
      }
    }
    cout << "\n";
  }
  return 0;
}