#include <bits/stdc++.h>
using namespace std;
const long long mod = (long long)1e9 + 7;
int t;
long long n, L, R;
long long a[500500], p[500500];
void solve() {
  cin >> n >> L >> R;
  a[1] = 2 * n - 2;
  for (int i = 2; i <= n - 1; ++i) a[i] = a[i - 1] - 2;
  a[n] = 1;
  for (int i = 1; i <= n; ++i) p[i] = p[i - 1] + a[i];
  long long l = 0, r = n, res = -1;
  while (l <= r) {
    long long mid = (l + r) / 2;
    if (p[mid] >= L) {
      r = mid - 1;
      res = mid;
    } else {
      l = mid + 1;
    }
  }
  long long cnt = R - L + 1;
  bool done = false;
  long long start = L - p[res - 1];
  if (res == n) {
    cout << 1 << "\n";
    return;
  }
  for (long long i = start; i <= 2 * (n - res); ++i) {
    if (cnt == 0) {
      done = true;
      break;
    }
    if (i % 2 == 1)
      cout << res << " ";
    else
      cout << res + i / 2 << " ";
    --cnt;
  }
  if (done) {
    cout << "\n";
    return;
  }
  for (long long level = res + 1; level <= n; ++level) {
    if (level == n) {
      cout << 1 << " ";
      cout << "\n";
      return;
    }
    for (long long i = 1; i <= 2 * (n - level); ++i) {
      if (i % 2 == 1)
        cout << level << " ";
      else
        cout << level + i / 2 << " ";
      --cnt;
      if (cnt == 0) {
        cout << "\n";
        return;
      }
    }
  }
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

In CPP, your task is to solve the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long mod = (long long)1e9 + 7;
int t;
long long n, L, R;
long long a[500500], p[500500];
void solve() {
  cin >> n >> L >> R;
  a[1] = 2 * n - 2;
  for (int i = 2; i <= n - 1; ++i) a[i] = a[i - 1] - 2;
  a[n] = 1;
  for (int i = 1; i <= n; ++i) p[i] = p[i - 1] + a[i];
  long long l = 0, r = n, res = -1;
  while (l <= r) {
    long long mid = (l + r) / 2;
    if (p[mid] >= L) {
      r = mid - 1;
      res = mid;
    } else {
      l = mid + 1;
    }
  }
  long long cnt = R - L + 1;
  bool done = false;
  long long start = L - p[res - 1];
  if (res == n) {
    cout << 1 << "\n";
    return;
  }
  for (long long i = start; i <= 2 * (n - res); ++i) {
    if (cnt == 0) {
      done = true;
      break;
    }
    if (i % 2 == 1)
      cout << res << " ";
    else
      cout << res + i / 2 << " ";
    --cnt;
  }
  if (done) {
    cout << "\n";
    return;
  }
  for (long long level = res + 1; level <= n; ++level) {
    if (level == n) {
      cout << 1 << " ";
      cout << "\n";
      return;
    }
    for (long long i = 1; i <= 2 * (n - level); ++i) {
      if (i % 2 == 1)
        cout << level << " ";
      else
        cout << level + i / 2 << " ";
      --cnt;
      if (cnt == 0) {
        cout << "\n";
        return;
      }
    }
  }
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) solve();
  return 0;
}

```