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 int mn = 1 * (int)(1e5) + 10;
const int mod = 1 * (int)(1e9) + 7;
const int mm = 1 * (int)(1e3) + 10;
const int base = 1 * (int)(1e9);
const bool aNs = 0;
int tt, ntest = 1;
void docfile() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  if (ifstream("test.inp")) {
    freopen("test.inp", "r", stdin);
    if (!aNs)
      freopen("test.out", "w", stdout);
    else
      freopen("test.ans", "w", stdout);
  } else if (ifstream("1334d.inp")) {
    freopen("1334d.inp", "r", stdin);
    freopen("1334d.out", "w", stdout);
  }
}
template <typename T>
void read(T& x) {
  x = 0;
  T f = 1;
  char ch = getchar();
  while (!isdigit(ch)) f = ch == '-' ? -f : f, ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  x *= f;
}
template <typename T>
void write(T a) {
  if (a < 0) {
    putchar('-');
    write(-a);
    return;
  }
  if (a < 10) {
    putchar('0' + a);
    return;
  }
  write(a / 10);
  putchar((char)('0' + (a % 10)));
}
void enter() {
  int n;
  long long l, r;
  cin >> n >> l >> r;
  long long s = 0;
  for (int i = 1; i < n; ++i) {
    if (s + 2 * (n - i) >= l) {
      vector<int> v;
      for (int j = i + 1; j <= n; ++j) {
        v.emplace_back(i);
        v.emplace_back(j);
      }
      for (int k = max(l - 1 - s, 0ll); k < min(r - s, 2ll * (n - i)); ++k)
        cout << v[k] << " ";
    }
    if (s + 2 * (n - i) >= r) {
      break;
    }
    s += 2 * (n - i);
  }
  if (r == 1ll * n * (n - 1) + 1) cout << 1;
}
void solve() { cout << "\n"; }
void print_result() {}
int main() {
  docfile();
  cin >> ntest;
  for (tt = 1; tt <= ntest; ++tt) {
    enter();
    solve();
    print_result();
  }
}