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;
void solve() {
  long long int i, j, n, m, k;
  long long int l, r;
  cin >> n >> l >> r;
  bool var1 = 0;
  long long int var2 = 0, var = 0;
  if (r == (n * (n - 1LL)) + 1LL) {
    var1 = 1;
    r--;
  } else {
    long long int o = 0;
    while (o < 2) o++;
  }
  long long int o = 0;
  for (i = 1; i <= n; i++) {
    var2 += (n - i) * 2LL;
    if (o == 1) {
      o = 0;
    } else if (var2 >= l) {
      var2 -= (n - i) * 2LL;
      var = i;
      break;
    } else {
      o = 0;
    }
  }
  var2 = l - var2;
  long long int p = var2;
  long long int var3 = 1;
  var2 = (var + var2 / 2 + var2 % 2);
  if (l <= r && (p % 2LL == 0)) {
    cout << var2 << " ";
    l++;
    var2++;
  } else {
    o = 0;
  }
  o = 0;
  while (l <= r && o == 0) {
    if (var2 > n) {
      var++;
      var2 = var + 1;
    }
    if (o == 1)
      o = 0;
    else
      o = 0;
    if ((var3 % 2) == 1)
      cout << var << " ";
    else {
      cout << var2 << " ";
      var2++;
    }
    l++;
    var3 = 1 - var3;
  }
  o = 1;
  if (var1) cout << o << " ";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int i, j, n, m, k;
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
}