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 mods = 998244353;
const int maxn = 1e5 + 10;
const int N = 1e5 + 10;
const int E = 2e5 + 10;
long long n, l, r;
long long k[maxn];
vector<int> ans;
int main() {
  int T;
  cin >> T;
  while (T--) {
    cin >> n >> l >> r;
    k[1] = 1;
    for (int i = 2; i <= (n); ++i) {
      k[i] = k[i - 1] + 2 * (n - i + 1);
    }
    long long bo = n;
    for (int i = 1; i <= (n); ++i) {
      if (k[i] > l) {
        bo = i - 1;
        break;
      }
    }
    if (bo >= n) {
      printf("1\n");
      continue;
    }
    long long pc = k[bo];
    long long tot = 1;
    long long tmp;
    ans.clear();
    while (pc <= r) {
      if (bo == n) {
        ans.push_back(1);
        break;
      }
      if (pc % 2 == 1)
        tmp = bo;
      else {
        tmp = bo + tot;
        tot++;
      }
      if (pc >= l) ans.push_back(tmp);
      pc++;
      if (tmp == n) {
        bo++;
        tot = 1;
      }
    }
    for (int i = 0; i <= (ans.size() - 1); ++i) {
      cout << ans[i] << " ";
    }
    cout << endl;
  }
}