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;
const double eps = 1e-6;
const int maxn = 3e5 + 100;
const int maxm = 2e6 + 100;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
int t;
long long n;
long long l, r;
long long f(long long x) { return (2ll * n - x - 1) * x / 2ll; }
void pt(long long x) { printf("%lld ", x); }
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%lld %lld %lld", &n, &l, &r);
    long long id = (l + 1) / 2;
    int ok = 0;
    if (r > n * (n - 1)) {
      ok = 1;
      r--;
    }
    if (l <= r) {
      long long L = 1, R = n - 1;
      long long res;
      while (L <= R) {
        long long mid = L + R >> 1;
        if (f(mid) >= id) {
          res = mid;
          R = mid - 1;
        } else
          L = mid + 1;
      }
      long long j = id - f(res - 1);
      while (l <= r) {
        if (l % 2 == 0) {
          pt(j + res);
          j++;
          l++;
        } else {
          pt(res);
          l++;
        }
        if (j > n - res) {
          j = 1;
          res++;
        }
      }
    }
    if (ok) printf("1");
    printf("\n");
  }
  return 0;
}