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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using namespace std;
inline int qpow(int b, int e, int m = 998244353) {
  int a = 1;
  for (; e; e >>= 1, b = (long long)b * b % m)
    if (e & 1) a = (long long)a * b % m;
  return a;
}
int n, m, q, k;
int a[300005], b[300005], c[300005];
const int pp[11] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31};
void solve() {
  long long n, l, r;
  cin >> n >> l >> r;
  vector<long long> v(n + 1);
  v[1] = 1;
  for (long long i = 2; i < n; i++) {
    v[i] = v[i - 1] + 2 * (n - (i - 1));
  }
  v[n] = v[n - 1] + 2;
  auto itr1 = upper_bound(v.begin(), v.end(), l);
  auto itr2 = upper_bound(v.begin(), v.end(), r);
  if (itr1 == v.end()) {
    cout << 1 << endl;
  } else {
    int in1 = -1;
    int in2 = -1;
    auto k1 = itr1;
    auto k2 = itr2;
    int st1;
    st1 = (--k1) - v.begin();
    int st2;
    st2 = k2 - v.begin();
    long long i = v[st1];
    while (i <= r && i < v[n]) {
      for (int j = 0; j < n - st1; j++) {
        if (i >= l && i <= r) cout << st1 << " ";
        i++;
        if (i >= l && i <= r) cout << st1 + j + 1 << " ";
        i++;
      }
      st1++;
    }
    if (r == v[n])
      cout << 1 << endl;
    else
      cout << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}