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;
long long v[100010];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long n, l, r;
    cin >> n >> l >> r;
    for (int i = 1; i < n; ++i) v[i] = 2 * (n - i);
    v[n] = 1;
    for (int i = 1; i <= n; ++i) v[i] += v[i - 1];
    int a = 0, b = n + 1;
    while (b - a > 1) {
      int mid = (a + b) / 2;
      if (v[mid] >= l)
        b = mid;
      else
        a = mid;
    }
    long long current = v[a];
    long long pos = a + 1;
    while (current < l) {
      current += 2;
      pos++;
    }
    bool flag = current != l;
    long long lim = 1ll * n * (n - 1) + 1;
    for (long long i = l; i <= min(lim - 1, r); ++i) {
      if (flag)
        cout << a + 1 << " ";
      else
        cout << pos++ << " ";
      if (pos > n) {
        a++;
        pos = a + 2;
      }
      flag ^= 1;
    }
    if (lim == r) cout << "1";
    cout << "\n";
  }
  return 0;
}