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;
long long l, r;
int n;
vector<long long> a;
int main() {
  int tt;
  scanf("%d", &tt);
  while (tt--) {
    a.clear();
    scanf("%d%lld%lld", &n, &l, &r);
    long long sum = 0, pos1 = 0;
    for (int i = 1; i <= n; i++) {
      if (1 + sum <= l && 1ll * (n - i) * 2 + sum >= l) {
        pos1 = i;
        break;
      } else
        sum = sum + 1ll * (n - i) * 2;
    }
    if (l == 1ll * n * (n - 1) + 1)
      a.push_back(1);
    else {
      while (l <= r) {
        if (l % 2)
          a.push_back(pos1);
        else {
          long long t = l - sum - 1;
          t = pos1 + (t + 1) / 2;
          a.push_back(t);
        }
        l++;
        if (l > r) break;
        if (l > (1ll * n - pos1) * 2 + sum) {
          sum = sum + (1ll * n - pos1) * 2;
          pos1++;
        }
        if (l == 1ll * n * (n - 1) + 1) {
          a.push_back(1);
          break;
        }
      }
    }
    int cnt = a.size();
    for (int i = 0; i < cnt - 1; i++) printf("%lld ", a[i]);
    printf("%lld\n", a[cnt - 1]);
  }
  return 0;
}