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>
using namespace std;
namespace io {
inline int rdi() {
  int v;
  cin >> v;
  return v;
}
inline long long rdll() {
  long long v;
  cin >> v;
  return v;
}
inline string rds() {
  string s;
  cin >> s;
  return s;
}
inline char rdc() {
  char c;
  cin >> c;
  return c;
}
template <class T>
void rdv(T beg_iter, T end_iter) {
  for (; beg_iter != end_iter; ++beg_iter) cin >> *beg_iter;
}
inline void wri(const int v) { cout << v; }
inline void wrll(const long long &v) { cout << v; }
inline void wrs(const string &s) { cout << s; }
inline void wrc(const char c) { cout << c; }
template <class T>
void wrv(T beg_iter, T end_iter, string separator = " ",
         string end_of_line = "\n") {
  for (; beg_iter != end_iter; ++beg_iter) {
    cout << *beg_iter;
    cout << separator;
  }
  cout << end_of_line;
}
}  // namespace io
using namespace io;
bool sortbysec(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return (a.second < b.second);
}
long long binpow(long long a, long long b, long long m) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
string path = "DLRU";
vector<int> di = {1, 0, 0, -1};
vector<int> dj = {0, -1, 1, 0};
void solve();
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  if (10 == 1) {
    t = 1;
  } else {
    cin >> t;
  }
  while (t--) solve();
  return 0;
}
void solve() {
  long long n, x, y;
  cin >> n >> x >> y;
  vector<long long> A(n);
  long long dif = y - x;
  long long a = -1, d = -1, a1, d1;
  bool ok = 0;
  int mx = INT_MAX;
  for (long long i = 0; i < n; ++i) {
    for (long long j = i + 1; j < n; ++j) {
      if (dif % (j - i) == 0) {
        int d1 = dif / (j - i);
        if (x - i * d1 > 0) {
          a1 = x - i * d1;
          if (a1 + (n - 1) * d1 < mx) {
            a = a1;
            d = d1;
            mx = a + (n - 1) * d;
          }
        }
      }
    }
  }
  for (long long i = 0; i < n; ++i) A[i] = a + i * d;
  wrv(A.begin(), A.end());
}