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;
using ll = long long;
using pll = pair<ll, ll>;
const int LM = 3e5 + 4;
ll N;
ll L, R;
ll part[LM];
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%lld%lld%lld", &N, &L, &R);
    bool e = 0;
    if (R == N * (N - 1) + 1) {
      e = 1;
      R--;
    }
    for (int i = 1; i <= N - 1; i++) part[i] = part[i - 1] + (N - i) * 2;
    for (int i = 1; i <= N; i++) {
      if (L <= part[i] && R > part[i - 1]) {
        int v, last = i;
        for (int j = 1; j <= (N - i) * 2; j++) {
          if (j & 1)
            v = i;
          else
            v = ++last;
          if (L <= part[i - 1] + j && part[i - 1] + j <= R) printf("%d ", v);
        }
      }
    }
    if (e) printf("1 ");
    puts("");
  }
  return 0;
}