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;
template <typename T>
T InvMod(T a, T b, T &x, T &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  T x1, y1;
  T g = InvMod(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return g;
}
long long fact(long long n) {
  if (n == 1) return 1;
  return (n % 1000000007 * fact(n - 1) % 1000000007) % 1000000007;
}
void sieve(vector<int> &v) {
  bool arr[1000004];
  memset(arr, true, sizeof arr);
  for (int i = 2; i * i <= 1000002; i++) {
    if (arr[i] == true) {
      v.push_back(i);
      for (int j = i * i; j <= 1000002; j += i) {
        arr[j] = false;
      }
    }
  }
}
bool cmp(pair<int, int> a, pair<int, int> b) { return a.first > b.first; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long n, l, r;
    cin >> n >> l >> r;
    int i = 1;
    long long tmp = l;
    for (long long j = l; j <= r; j++) {
      while (2 * (n - i) < tmp && i <= n) {
        tmp -= 2 * (n - i);
        i++;
      }
      if (i > n) {
        cout << 1 << " ";
        continue;
      }
      long long rem = tmp;
      if (rem % 2 == 1) {
        cout << i << " ";
      } else {
        cout << i + (rem / 2) << " ";
      }
      tmp++;
    }
    cout << endl;
  }
}