#include <bits/stdc++.h>
using ll = long long;
using namespace std;
const ll N = 2e5 + 5, mod = 1e9 + 7;
const ll inf = 1e18;
struct cmp {
  bool operator()(pair<int, int> a, pair<int, int> b) {
    if (a.first == b.first) return a.second < b.second;
    return a.first > b.first;
  }
};
ll power(ll x, ll p) {
  ll r = 1ll;
  x = x % mod;
  while (p > 0) {
    if (p & 1) r = (r * x) % mod;
    p = p >> 1;
    x = (x * x) % mod;
  }
  return r;
}
void solve() {
  ll n, l, r;
  cin >> n >> l >> r;
  ll firstrow = n + (n - 2);
  ll ele = 0, i = 0;
  bool edge = 0;
  while (ele < l) {
    if (firstrow - 2ll * i > 0)
      ele += firstrow - 2ll * i;
    else {
      ele++;
      edge = 1;
    }
    i++;
  }
  if (!edge) ele -= firstrow - 2ll * (i - 1);
  bool flag = 0;
  ll j = 1;
  if (!edge) ele++;
  while (ele <= r) {
    ll ans;
    if (flag == 0)
      ans = i;
    else {
      ans = i + j;
      j++;
    }
    flag = 1 - flag;
    if (ele >= l && !edge) cout << ans << " ";
    if (edge) {
      cout << 1 << endl;
      return;
    }
    if (i + j - 1 >= n) {
      i++;
      j = 1;
      flag = 0;
      if (i >= n) i = 1;
    }
    ele++;
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1, p;
  cin >> t;
  for (p = 1; p <= t; p++) {
    solve();
  }
  return 0;
}

### Prompt

Please create a solution in CPP to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using ll = long long;
using namespace std;
const ll N = 2e5 + 5, mod = 1e9 + 7;
const ll inf = 1e18;
struct cmp {
  bool operator()(pair<int, int> a, pair<int, int> b) {
    if (a.first == b.first) return a.second < b.second;
    return a.first > b.first;
  }
};
ll power(ll x, ll p) {
  ll r = 1ll;
  x = x % mod;
  while (p > 0) {
    if (p & 1) r = (r * x) % mod;
    p = p >> 1;
    x = (x * x) % mod;
  }
  return r;
}
void solve() {
  ll n, l, r;
  cin >> n >> l >> r;
  ll firstrow = n + (n - 2);
  ll ele = 0, i = 0;
  bool edge = 0;
  while (ele < l) {
    if (firstrow - 2ll * i > 0)
      ele += firstrow - 2ll * i;
    else {
      ele++;
      edge = 1;
    }
    i++;
  }
  if (!edge) ele -= firstrow - 2ll * (i - 1);
  bool flag = 0;
  ll j = 1;
  if (!edge) ele++;
  while (ele <= r) {
    ll ans;
    if (flag == 0)
      ans = i;
    else {
      ans = i + j;
      j++;
    }
    flag = 1 - flag;
    if (ele >= l && !edge) cout << ans << " ";
    if (edge) {
      cout << 1 << endl;
      return;
    }
    if (i + j - 1 >= n) {
      i++;
      j = 1;
      flag = 0;
      if (i >= n) i = 1;
    }
    ele++;
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1, p;
  cin >> t;
  for (p = 1; p <= t; p++) {
    solve();
  }
  return 0;
}

```