#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5;
const double pi = acos(-1.0);
const long long int inf = 0x3f3f3f3f3f3f3f3f;
const long long int mod = 998244353;
bool isPowerOfTwo(int x) { return x && (!(x & (x - 1))); }
void fast() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int modInverse(long long int n, long long int p) {
  return power(n, p - 2, p);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, l, r;
    cin >> n >> l >> r;
    long long int k1 = 0, k2 = 0, i1, i2;
    for (i2 = 1; i2 < n; i2++) {
      k2 += 2 * (n - i2);
      if (k2 >= r) break;
    }
    for (i1 = 1; i1 < n; i1++) {
      k1 += 2 * (n - i1);
      if (k1 >= l) break;
    }
    i2++;
    vector<long long int> v, a;
    k1 -= (n - i1) * 2;
    for (int i = i1; i < i2; i++) {
      for (int j = i + 1; j < n + 1; j++) {
        a.emplace_back(i);
        a.emplace_back(j);
      }
    }
    for (long long int i = l; i <= r; i++) {
      if (i == (n * (n - 1) + 1))
        cout << "1 ";
      else
        cout << a[i - k1 - 1] << " ";
    }
    cout << '\n';
  }
  return 0;
}

### 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;
const int N = 3e5 + 5;
const double pi = acos(-1.0);
const long long int inf = 0x3f3f3f3f3f3f3f3f;
const long long int mod = 998244353;
bool isPowerOfTwo(int x) { return x && (!(x & (x - 1))); }
void fast() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int modInverse(long long int n, long long int p) {
  return power(n, p - 2, p);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, l, r;
    cin >> n >> l >> r;
    long long int k1 = 0, k2 = 0, i1, i2;
    for (i2 = 1; i2 < n; i2++) {
      k2 += 2 * (n - i2);
      if (k2 >= r) break;
    }
    for (i1 = 1; i1 < n; i1++) {
      k1 += 2 * (n - i1);
      if (k1 >= l) break;
    }
    i2++;
    vector<long long int> v, a;
    k1 -= (n - i1) * 2;
    for (int i = i1; i < i2; i++) {
      for (int j = i + 1; j < n + 1; j++) {
        a.emplace_back(i);
        a.emplace_back(j);
      }
    }
    for (long long int i = l; i <= r; i++) {
      if (i == (n * (n - 1) + 1))
        cout << "1 ";
      else
        cout << a[i - k1 - 1] << " ";
    }
    cout << '\n';
  }
  return 0;
}

```