#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, pstart, pend;
  int t;
  cin >> t;
  for (int testcases = 1; testcases <= t; testcases++) {
    cin >> n >> pstart >> pend;
    bool completestart = true;
    bool completeend = true;
    bool flag = false;
    bool paranoidcheck = false;
    if (pstart == pend) {
      paranoidcheck = true;
    }
    if (pend == n * (n - 1) + 1) {
      if (pstart == pend) {
        cout << 1 << endl;
        continue;
      } else {
        flag = true;
        pend -= 1;
      }
    }
    if (pstart % 2 == 0) {
      completestart = false;
    } else {
      pstart += 1;
    }
    pstart /= 2;
    if (pend % 2 == 1) {
      completeend = false;
      pend += 1;
    }
    pend /= 2;
    long long iter = pend - pstart;
    if (paranoidcheck) {
      iter -= 1;
    }
    long long a = 1;
    long long b = 1;
    while (pstart - (n - a) > 0) {
      pstart -= (n - a);
      a += 1;
    }
    b = a + pstart;
    if (!completestart) {
      cout << b << " ";
      if (b < n) {
        b += 1;
      } else {
        a += 1;
        b = a + 1;
      }
      iter -= 1;
    }
    if (!completeend) {
      iter -= 1;
    }
    for (int k = 0; k <= iter; k++) {
      cout << a << " " << b << " ";
      if (b < n) {
        b += 1;
      } else {
        a += 1;
        b = a + 1;
      }
    }
    if (!completeend) {
      cout << a << " ";
    }
    if (flag) {
      cout << 1;
    }
    cout << endl;
  }
  return 0;
}

### Prompt

Please formulate a CPP solution to 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>
#pragma GCC optimize("Ofast")
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, pstart, pend;
  int t;
  cin >> t;
  for (int testcases = 1; testcases <= t; testcases++) {
    cin >> n >> pstart >> pend;
    bool completestart = true;
    bool completeend = true;
    bool flag = false;
    bool paranoidcheck = false;
    if (pstart == pend) {
      paranoidcheck = true;
    }
    if (pend == n * (n - 1) + 1) {
      if (pstart == pend) {
        cout << 1 << endl;
        continue;
      } else {
        flag = true;
        pend -= 1;
      }
    }
    if (pstart % 2 == 0) {
      completestart = false;
    } else {
      pstart += 1;
    }
    pstart /= 2;
    if (pend % 2 == 1) {
      completeend = false;
      pend += 1;
    }
    pend /= 2;
    long long iter = pend - pstart;
    if (paranoidcheck) {
      iter -= 1;
    }
    long long a = 1;
    long long b = 1;
    while (pstart - (n - a) > 0) {
      pstart -= (n - a);
      a += 1;
    }
    b = a + pstart;
    if (!completestart) {
      cout << b << " ";
      if (b < n) {
        b += 1;
      } else {
        a += 1;
        b = a + 1;
      }
      iter -= 1;
    }
    if (!completeend) {
      iter -= 1;
    }
    for (int k = 0; k <= iter; k++) {
      cout << a << " " << b << " ";
      if (b < n) {
        b += 1;
      } else {
        a += 1;
        b = a + 1;
      }
    }
    if (!completeend) {
      cout << a << " ";
    }
    if (flag) {
      cout << 1;
    }
    cout << endl;
  }
  return 0;
}

```