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;
using ll = long long;
using ld = long double;
template <typename T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  } else
    return false;
}
template <typename T>
bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return true;
  } else
    return false;
}
ll S(ll n, ll k) { return 2 * (n - k); }
void solve() {
  int T;
  cin >> T;
  for (ll(q) = (1); (q) <= (T); (q)++) {
    ll n, l, r;
    cin >> n >> l >> r;
    ll length = r - l + 1;
    ll lk;
    for (ll(k) = (1); (k) <= (n - 1); (k)++) {
      if (l - S(n, k) > 0)
        l -= S(n, k), r -= S(n, k);
      else {
        lk = k;
        break;
      }
    }
    vector<ll> res;
    {
      res.push_back(0);
      ll i = 0;
      for (ll(k) = (lk); (k) <= (n - 1); (k)++) {
        for (ll(j) = (k + 1); (j) <= (n); (j)++) {
          res.push_back(k);
          i++;
          if (i == r) goto hoge;
          res.push_back(j);
          i++;
          if (i == r) goto hoge;
        }
      }
      if (i < r) res.push_back(1), i++;
    hoge:
      assert(i == r);
    }
    for (ll(i) = (l); (i) <= (r); (i)++) {
      cout << res[i] << (i < r ? " " : "\n");
    }
  }
  return;
}
int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  solve();
  return 0;
}