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 n;
void prnt(long long a, long long b, long long cnt) {
  if (cnt) printf("%d ", a);
  if (cnt < 2) return;
  printf("%d ", b);
  if (b != n)
    prnt(a, b + 1, cnt - 2);
  else if (a != n - 1)
    prnt(a + 1, a + 2, cnt - 2);
  else if (cnt > 2)
    printf("1");
}
long long arth(long long st, long long en, long long n) {
  return n * (st + en) / 2;
}
long long clc(long long m) { return arth((n - 1) * 2, (n - m) * 2, m); }
long long bs(long long lft) {
  long long l = 1, r = n - 1, m;
  while (l < r) {
    m = (l + r) / 2;
    long long c = clc(m);
    if (lft > c)
      l = m + 1;
    else
      r = m;
  }
  return l;
}
void fnd(long long lft, long long rit) {
  long long l1 = bs(lft);
  long long l2 = bs(lft + 1);
  if (l1 == l2) {
    long long nl = lft;
    if (l1 > 1) nl -= clc(l1 - 1);
    long long k = nl / 2;
    if (nl % 2 == 0) cout << k + l1 << " ", lft++;
    prnt(l1, k + l1 + 1, rit - lft + 1);
  } else {
    cout << n << " ";
    prnt(l2, l2 + 1, rit - lft);
  }
}
int main() {
  long long t;
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld", &n);
    long long lft, rit;
    scanf("%lld%lld", &lft, &rit);
    long long mx = n * (n - 1) + 1;
    if (lft == mx)
      cout << 1 << endl;
    else if (lft + 1 == mx) {
      cout << n;
      if (rit == mx) cout << " " << 1 << endl;
    } else {
      fnd(lft, rit);
      cout << endl;
    }
  }
}