Problem: We have a secret array. You don't know this array and you have to restore it. However, you know some facts about this array:

  * The array consists of n distinct positive (greater than 0) integers. 
  * The array contains two elements x and y (these elements are known for you) such that x < y. 
  * If you sort the array in increasing order (such that a_1 < a_2 < … < a_n), differences between all adjacent (consecutive) elements are equal (i.e. a_2 - a_1 = a_3 - a_2 = … = a_n - a_{n-1}). 



It can be proven that such an array always exists under the constraints given below.

Among all possible arrays that satisfy the given conditions, we ask you to restore one which has the minimum possible maximum element. In other words, you have to minimize max(a_1, a_2, ..., a_n).

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.

The only line of the test case contains three integers n, x and y (2 ≤ n ≤ 50; 1 ≤ x < y ≤ 50) — the length of the array and two elements that are present in the array, respectively.

Output

For each test case, print the answer: n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of the required array. If there are several answers, you can print any (it also means that the order of elements doesn't matter).

It can be proven that such an array always exists under the given constraints.

Example

Input


5
2 1 49
5 20 50
6 20 50
5 3 8
9 13 22


Output


1 49 
20 40 30 50 10
26 32 20 38 44 50 
8 23 18 13 3 
1 10 13 4 19 22 25 16 7 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC target("avx2,avx")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
using ll = long long;
using ull = unsigned long long;
using i128 = __int128_t;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
template <class T>
using vec = vector<T>;
template <class T>
using vvec = vector<vector<T>>;
constexpr char ln = '\n';
istream& operator>>(istream& is, __int128_t& x) {
  x = 0;
  string s;
  is >> s;
  int n = int(s.size()), it = 0;
  if (s[0] == '-') it++;
  for (; it < n; it++) x = (x * 10 + s[it] - '0');
  if (s[0] == '-') x = -x;
  return is;
}
ostream& operator<<(ostream& os, __int128_t x) {
  if (x == 0) return os << 0;
  if (x < 0) os << '-', x = -x;
  deque<int> deq;
  while (x) deq.emplace_front(x % 10), x /= 10;
  for (int e : deq) os << e;
  return os;
}
template <class T1, class T2>
ostream& operator<<(ostream& os, const pair<T1, T2>& p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "{";
  for (int i = 0; i < int(v.size()); i++) {
    if (i) os << ", ";
    os << v[i];
  }
  return os << "}";
}
template <class Container>
inline int SZ(Container& v) {
  return int(v.size());
}
template <class T>
inline void UNIQUE(vector<T>& v) {
  v.erase(unique(v.begin(), v.end()), v.end());
}
template <class T1, class T2>
inline bool chmax(T1& a, T2 b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T1, class T2>
inline bool chmin(T1& a, T2 b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
inline int topbit(ull x) { return x == 0 ? -1 : 63 - __builtin_clzll(x); }
inline int botbit(ull x) { return x == 0 ? 64 : __builtin_ctzll(x); }
inline int popcount(ull x) { return __builtin_popcountll(x); }
inline int kthbit(ull x, int k) { return (x >> k) & 1; }
inline constexpr long long TEN(int x) { return x == 0 ? 1 : TEN(x - 1) * 10; }
inline void print() { cout << "\n"; }
template <class T>
inline void print(const vector<T>& v) {
  for (int i = 0; i < int(v.size()); i++) {
    if (i) cout << " ";
    cout << v[i];
  }
  print();
}
template <class T, class... Args>
inline void print(const T& x, const Args&... args) {
  cout << x << " ";
  print(args...);
}
struct fast_ios {
  fast_ios() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(20);
  };
} fast_ios_;
void solve() {
  int n, x, y;
  cin >> n >> x >> y;
  vec<int> ans;
  for (int d = 1; d <= y - x; d++) {
    if ((y - x) % d != 0) continue;
    deque<int> deq;
    for (int v = x; v <= y; v += d) {
      deq.emplace_back(v);
    }
    if (SZ(deq) > n) continue;
    while (SZ(deq) < n) {
      int v = deq.front() - d;
      if (v <= 0) break;
      deq.emplace_front(v);
    }
    while (SZ(deq) < n) {
      int v = deq.back() + d;
      deq.emplace_back(v);
    }
    if (ans.empty() or ans.back() > deq.back()) {
      ans.clear();
      while (!deq.empty()) {
        ans.emplace_back(deq.front());
        deq.pop_front();
      }
    }
  }
  print(ans);
}
int main() {
  int Q;
  cin >> Q;
  while (Q--) {
    solve();
  }
}