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 cnt[100006];
vector<int> ans;
int main() {
  int tests = 1;
  int n;
  long long l, r;
  scanf("%d", &tests);
  while (tests--) {
    ans.clear();
    scanf("%d%lld%lld", &n, &l, &r);
    for (int i = 1; i < n; ++i) cnt[i] = 1ll * (2 * n - i - 1) * i;
    int s = 0, t = 0, f = 0;
    if (r == 1ll * n * (n - 1) + 1) f = 1;
    for (int i = 1; i < n; ++i) {
      if (cnt[i] < l) s = i;
      if (cnt[i] <= r) t = i;
    }
    s++;
    t++;
    for (int i = s; i <= t; ++i) {
      for (int j = i + 1; j <= n; ++j) ans.push_back(i), ans.push_back(j);
    }
    r -= cnt[s - 1] + 1, l -= cnt[s - 1] + 1;
    for (int i = 0; i < (int)ans.size(); ++i) {
      if (i >= l && i <= r) printf("%d ", ans[i]);
    }
    if (f) printf("1");
    puts("");
  }
}