#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    ll l, r;
    cin >> l >> r;
    ll idx = 1;
    bool done = false;
    for (int start = 1; start <= n; start++) {
      if (idx <= l && idx + (n - start) * 2ll > l) {
        int now = start;
        int nxt = start + 1;
        for (; idx < l; idx++) {
          int prev_now = now;
          now = nxt;
          nxt = (prev_now == start ? start : prev_now + 1);
          if (nxt == n + 1) {
            assert(now == start);
            now = ++start;
            nxt = start + 1;
          }
        }
        for (; idx <= r; idx++) {
          cout << now << " ";
          int prev_now = now;
          now = nxt;
          nxt = (prev_now == start ? start : prev_now + 1);
          if (nxt == n + 1) {
            assert(now == start);
            now = ++start;
            nxt = start + 1;
            if (now == n) {
              now = 1;
            }
          }
        }
        cout << endl;
        done = true;
        break;
      } else {
        idx += (n - start) * 2ll;
      }
    }
    if (!done) cout << 1 << '\n';
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    ll l, r;
    cin >> l >> r;
    ll idx = 1;
    bool done = false;
    for (int start = 1; start <= n; start++) {
      if (idx <= l && idx + (n - start) * 2ll > l) {
        int now = start;
        int nxt = start + 1;
        for (; idx < l; idx++) {
          int prev_now = now;
          now = nxt;
          nxt = (prev_now == start ? start : prev_now + 1);
          if (nxt == n + 1) {
            assert(now == start);
            now = ++start;
            nxt = start + 1;
          }
        }
        for (; idx <= r; idx++) {
          cout << now << " ";
          int prev_now = now;
          now = nxt;
          nxt = (prev_now == start ? start : prev_now + 1);
          if (nxt == n + 1) {
            assert(now == start);
            now = ++start;
            nxt = start + 1;
            if (now == n) {
              now = 1;
            }
          }
        }
        cout << endl;
        done = true;
        break;
      } else {
        idx += (n - start) * 2ll;
      }
    }
    if (!done) cout << 1 << '\n';
  }
}

```