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;
inline long long read() {
  register char c = ' ';
  register long long v = 0, x = 1;
  while ((c < '0') || (c > '9')) {
    if (c == '-') x = -1;
    c = getchar();
  }
  while ((c >= '0') && (c <= '9')) {
    v = (v << 1) + (v << 3) + (c ^ 48);
    c = getchar();
  }
  return v * x;
}
inline void write(register long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 ^ 48);
}
long long t, n, l, r, x, y;
inline void calc(register long long m) {
  if (m > n * (n - 1) / 2) return (void)(x = y = 1ll);
  register long long a = 0, b = 0;
  while (a < m) a += n - 1 - b, b++;
  a -= n - b, x = b, y = m - a + b;
}
inline void nxt() {
  (y != n) ? (y++) : (x++, y = x + 1);
  if (x == n) x = 1;
}
int main() {
  t = read();
  while (t--) {
    n = read(), l = read(), r = read(), calc((l + 1) >> 1);
    for (register long long i = l; i <= r; i++, putchar(' '))
      (i & 1) ? write(x) : (write(y), nxt());
    putchar('\n');
  }
  return ~~(0 - 0);
}