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 maxn = (int)1e6 + 100;
const int mod = (int)1e9 + 7;
void solve() {
  long long n, l, r;
  scanf("%lld%lld%lld", &n, &l, &r);
  long long pos = -1, pre = 0;
  for (auto i = (1); i <= (n); ++i) {
    long long now = (n - i) * 2;
    if (now == 0) now = 1;
    if (pre < l && pre + now >= l) {
      pos = i;
      break;
    }
    pre += now;
  }
  long long num = r - l + 1, st = l - pre, fr = pos == n ? 1 : pos,
            se = (st + 1) / 2 + pos;
  for (auto i = (1); i <= (num); ++i) {
    if (st % 2)
      printf("%lld ", fr > n ? 1 : fr);
    else
      printf("%lld ", se > n ? 1 : se), se++;
    st++;
    if (se > n) {
      pos++;
      st = 1;
      fr = pos;
      se = pos + 1;
      if (fr == n) fr = 1;
      if (se > n) se = 1;
    }
  }
  puts("");
}
int main() {
  int T;
  cin >> T;
  while (T--) solve();
}