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;
char buf[1 << 21], *p1 = buf, *p2 = buf, obuf[1 << 23], *O = obuf;
inline int read() {
  int x = 0, sign = 0;
  char s = getchar();
  while (!isdigit(s)) sign |= s == '-', s = getchar();
  while (isdigit(s)) x = (x << 1) + (x << 3) + (s - '0'), s = getchar();
  return sign ? -x : x;
}
void print(int x) {
  if (x > 9) print(x / 10);
  (*O++ = x % 10 + '0');
}
const int N = 3e5 + 5;
long long n, l, r;
void solve() {
  cin >> n >> l >> r;
  long long cnt = n, now = 1, tmp = 1;
  while (--cnt) {
    long long nw = tmp + cnt * 2 - 1;
    long long nl = max(l, tmp), nr = min(nw, r);
    if (nl <= nr)
      for (long long i = nl; i <= nr; i++) {
        if ((i - tmp) & 1)
          cout << now + (i - tmp + 1) / 2 << " ";
        else
          cout << now << " ";
      }
    tmp = nw + 1, now++;
  }
  if (r == tmp) cout << "1 ";
  puts("");
}
int main() {
  int testcase = 1;
  cin >> testcase;
  while (testcase--) solve();
  return 0;
}