#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
using vi = vector<ll>;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ll t;
  cin >> t;
  while (t--) {
    ll n, l, r;
    cin >> n >> l >> r;
    ll sm = 0;
    ll iters = 0;
    ll add = n - 1;
    if (l == n * (n - 1) + 1) {
      cout << 1 << endl;
      continue;
    }
    while (2 * sm < l) {
      sm += add;
      add--;
      iters++;
    }
    sm -= (add + 1);
    sm *= 2;
    sm++;
    ll currnext = iters + 1;
    bool startonone = true;
    while (sm < l) {
      if (sm + 1 < l) {
        sm += 2;
        currnext++;
      } else {
        startonone = false;
        sm++;
      }
    }
    for (ll i = l; i <= min(r, n * (n - 1)); i++) {
      ll toprint;
      if (startonone) {
        if ((i - l) % 2 == 0)
          toprint = iters;
        else {
          toprint = currnext;
          currnext++;
        }
      } else {
        if ((i - l) % 2 == 0) {
          toprint = currnext;
          currnext++;
        } else {
          toprint = iters;
        }
      }
      if (toprint == n) {
        iters++;
        currnext = iters + 1;
      }
      cout << toprint << ' ';
    }
    if (r == n * (n - 1) + 1) {
      cout << 1 << ' ';
    }
    cout << endl;
  }
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
using pii = pair<ll, ll>;
using vi = vector<ll>;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ll t;
  cin >> t;
  while (t--) {
    ll n, l, r;
    cin >> n >> l >> r;
    ll sm = 0;
    ll iters = 0;
    ll add = n - 1;
    if (l == n * (n - 1) + 1) {
      cout << 1 << endl;
      continue;
    }
    while (2 * sm < l) {
      sm += add;
      add--;
      iters++;
    }
    sm -= (add + 1);
    sm *= 2;
    sm++;
    ll currnext = iters + 1;
    bool startonone = true;
    while (sm < l) {
      if (sm + 1 < l) {
        sm += 2;
        currnext++;
      } else {
        startonone = false;
        sm++;
      }
    }
    for (ll i = l; i <= min(r, n * (n - 1)); i++) {
      ll toprint;
      if (startonone) {
        if ((i - l) % 2 == 0)
          toprint = iters;
        else {
          toprint = currnext;
          currnext++;
        }
      } else {
        if ((i - l) % 2 == 0) {
          toprint = currnext;
          currnext++;
        } else {
          toprint = iters;
        }
      }
      if (toprint == n) {
        iters++;
        currnext = iters + 1;
      }
      cout << toprint << ' ';
    }
    if (r == n * (n - 1) + 1) {
      cout << 1 << ' ';
    }
    cout << endl;
  }
}

```