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>
const long long int maxn = 1e10;
const long long mod = 1e9 + 7;
using namespace std;
long long int power(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b % 2 == 1) res = res * a;
    a = a * a;
    b = b / 2;
  }
  return res;
}
long long po(long long x, long long y) {
  long long res = 1;
  x = x % mod;
  while (y > 0) {
    if (y & 1) res = ((res % mod) * (x % mod)) % mod;
    y = y >> 1;
    x = ((x % mod) * (x % mod)) % mod;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    long long int n, l, r;
    cin >> n >> l >> r;
    long long int count = 0;
    int i = 1;
    long long int ini = 2 * (n - 1);
    while (1) {
      if (count + ini >= l || i == n) break;
      count += ini;
      ini -= 2;
      i++;
    }
    count++;
    vector<pair<long long int, long long int> > p;
    while (count <= r) {
      if (count == n * (n - 1) + 1) {
        p.push_back({1, count});
        count++;
        continue;
      }
      int it = i + 1, flag = 1;
      for (int j = 0; j < ini; j++) {
        if (flag) {
          p.push_back({i, count});
          count++;
          flag = 0;
        } else {
          p.push_back({it, count});
          count++;
          it++;
          flag = 1;
        }
      }
      ini -= 2;
      i++;
    }
    for (int j = 0; j < p.size(); j++) {
      if (p[j].second <= r && p[j].second >= l) cout << p[j].first << " ";
    }
    cout << endl;
  }
}