#include <bits/stdc++.h>
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native,avx,avx2,fma")
using namespace std;
vector<long long> v;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long a = 0, b = 0, c, d, e, f = 0, l, g, m, n, k, i, j, t, p, q, r, lo,
            hi, mid;
  cin >> t;
  while (t--) {
    cin >> n >> l >> r;
    if (l == n * (n - 1) + 1) {
      cout << 1 << '\n';
      continue;
    }
    lo = 1;
    hi = n;
    while (lo <= hi) {
      mid = (lo + hi) / 2;
      q = n - mid;
      p = q * (q + 1) + 1;
      p = n * (n - 1) + 1 - p;
      if (p >= l) {
        hi = mid - 1;
      } else {
        k = mid;
        lo = mid + 1;
      }
    }
    g = r - l + 1;
    q = n - k;
    p = q * (q + 1) + 1;
    p = n * (n - 1) + 1 - p;
    d = l - p;
    d = ceil(1.0 * d / 2.0);
    d += k;
    if (l % 2 == 0) {
      cout << d << ' ';
      if (d == n) {
        k++;
        d = k + 1;
      } else {
        d++;
      }
      l++;
    }
    for (i = l; i <= r; i++) {
      if (i % 2) {
        if (k == n) {
          cout << 1 << ' ';
        } else
          cout << k << ' ';
      } else {
        cout << d << ' ';
        if (d == n) {
          k++;
          d = k + 1;
        } else {
          d++;
        }
      }
    }
    cout << '\n';
  }
  return 0;
}

### Prompt

Develop a solution in cpp 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>
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native,avx,avx2,fma")
using namespace std;
vector<long long> v;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long a = 0, b = 0, c, d, e, f = 0, l, g, m, n, k, i, j, t, p, q, r, lo,
            hi, mid;
  cin >> t;
  while (t--) {
    cin >> n >> l >> r;
    if (l == n * (n - 1) + 1) {
      cout << 1 << '\n';
      continue;
    }
    lo = 1;
    hi = n;
    while (lo <= hi) {
      mid = (lo + hi) / 2;
      q = n - mid;
      p = q * (q + 1) + 1;
      p = n * (n - 1) + 1 - p;
      if (p >= l) {
        hi = mid - 1;
      } else {
        k = mid;
        lo = mid + 1;
      }
    }
    g = r - l + 1;
    q = n - k;
    p = q * (q + 1) + 1;
    p = n * (n - 1) + 1 - p;
    d = l - p;
    d = ceil(1.0 * d / 2.0);
    d += k;
    if (l % 2 == 0) {
      cout << d << ' ';
      if (d == n) {
        k++;
        d = k + 1;
      } else {
        d++;
      }
      l++;
    }
    for (i = l; i <= r; i++) {
      if (i % 2) {
        if (k == n) {
          cout << 1 << ' ';
        } else
          cout << k << ' ';
      } else {
        cout << d << ' ';
        if (d == n) {
          k++;
          d = k + 1;
        } else {
          d++;
        }
      }
    }
    cout << '\n';
  }
  return 0;
}

```