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;
void print_pattern(long long row, long long col, long long count,
                   long long elements) {
  if (elements == 0) {
    cout << 1 << " ";
    return;
  }
  long long x = row;
  long long next_x = row + 1;
  long long j = 0;
  bool flag = 1;
  while (j < elements && count > 0) {
    if (j >= col) {
      cout << x << " ";
      count--;
    }
    if (flag) {
      x = next_x;
      next_x++;
    } else {
      x = row;
    }
    flag = !flag;
    j++;
  }
  if (count > 0) {
    print_pattern(row + 1, 0, count, elements - 2);
  }
}
void solve() {
  long long n, l, r, count;
  cin >> n >> l >> r;
  count = r - l + 1;
  bool zero_flag = 0;
  long long i, j, row;
  long long to_subtract = 2 * (n - 1);
  row = 1;
  while (l - to_subtract > 0) {
    l -= to_subtract;
    to_subtract -= 2;
    row++;
    if (to_subtract == 0) {
      zero_flag = 1;
      break;
    }
  }
  if (zero_flag) {
    cout << 1 << endl;
    return;
  }
  if (l == 0) {
    row++;
  } else {
    l--;
  }
  print_pattern(row, l, count, to_subtract);
  cout << endl;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}