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 long long N = 300010;
long long t, n, a[N], b[N], x[N];
int main() {
  scanf("%lld", &t);
  while (t--) {
    long long l, r;
    scanf("%lld", &n);
    scanf("%lld%lld", &l, &r);
    long long tot = r - l + 1;
    long long high = n - 1, low = 1, mid, s = -1;
    while (high >= low) {
      mid = (high + low) / 2;
      long long c = (2 * ((mid * n) - (mid * (mid + 1)) / 2));
      if (c >= l) {
        s = mid;
        high = mid - 1;
      } else {
        low = mid + 1;
      }
    }
    if (s == -1) {
      printf("1\n");
      continue;
    }
    long long ps = s - 1, pre = (2 * ((ps * n) - (ps * (ps + 1)) / 2));
    long long pos = l - pre;
    while (tot > 0 && s < n) {
      long long len = (2 * (n - s));
      while (tot > 0 && pos <= len) {
        if (pos % 2)
          printf("%lld ", s);
        else {
          printf("%lld ", s + (pos / 2));
        }
        pos++;
        tot--;
      }
      if (tot > 0) {
        pos = 1;
        s++;
      }
    }
    if (tot) printf("1 ");
    printf("\n");
  }
  return 0;
}