#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")
int T;
int n, L;
long long l, r;
vector<int> v = {1, 2, 1};
long long cyc(int x) {
  if (x == n - 1) return 3;
  long long res = 2 * (n - x);
  return res;
}
void printcyc(int nr, long long start, int len) {
  if (nr == n - 1) {
    int step = 1;
    int x = n - 1;
    while (len > 0) {
      if (start > 0)
        start--;
      else {
        printf("%d ", x);
        len--;
      }
      if (step == 1) x = n;
      if (step == 2) x = 1;
      step++;
    }
    return;
  }
  int loops = start / 2;
  int c = nr + 1 + loops;
  int x = c;
  if (start % 2 == 0) x = nr;
  while (len > 0) {
    printf("%d ", x);
    len--;
    if (x != nr) {
      if (c == n) {
        printcyc(nr + 1, 0, len);
        break;
      } else {
        c++;
        x = nr;
      }
    } else {
      x = c;
    }
  }
}
void solve() {
  long long t = 1;
  for (int i = 1; i <= n - 1; i++) {
    if (t + cyc(i) < l)
      t += cyc(i);
    else {
      printcyc(i, l - t, L);
      break;
    }
  }
  printf("\n");
}
int main() {
  scanf("%d", &T);
  for (int test = 1; test <= T; test++) {
    scanf("%d%lld%lld", &n, &l, &r);
    if (n == 2) {
      for (int i = l - 1; i < r; i++) printf("%d ", v[i]);
      printf("\n");
      continue;
    }
    L = r - l + 1;
    solve();
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")
int T;
int n, L;
long long l, r;
vector<int> v = {1, 2, 1};
long long cyc(int x) {
  if (x == n - 1) return 3;
  long long res = 2 * (n - x);
  return res;
}
void printcyc(int nr, long long start, int len) {
  if (nr == n - 1) {
    int step = 1;
    int x = n - 1;
    while (len > 0) {
      if (start > 0)
        start--;
      else {
        printf("%d ", x);
        len--;
      }
      if (step == 1) x = n;
      if (step == 2) x = 1;
      step++;
    }
    return;
  }
  int loops = start / 2;
  int c = nr + 1 + loops;
  int x = c;
  if (start % 2 == 0) x = nr;
  while (len > 0) {
    printf("%d ", x);
    len--;
    if (x != nr) {
      if (c == n) {
        printcyc(nr + 1, 0, len);
        break;
      } else {
        c++;
        x = nr;
      }
    } else {
      x = c;
    }
  }
}
void solve() {
  long long t = 1;
  for (int i = 1; i <= n - 1; i++) {
    if (t + cyc(i) < l)
      t += cyc(i);
    else {
      printcyc(i, l - t, L);
      break;
    }
  }
  printf("\n");
}
int main() {
  scanf("%d", &T);
  for (int test = 1; test <= T; test++) {
    scanf("%d%lld%lld", &n, &l, &r);
    if (n == 2) {
      for (int i = l - 1; i < r; i++) printf("%d ", v[i]);
      printf("\n");
      continue;
    }
    L = r - l + 1;
    solve();
  }
}

```