#include<bits/stdc++.h>
using namespace std;
typedef  long long ll;
#define fat(x) for(auto it:x)
#define  pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define mem(x) memset(x,0,sizeof(x))
#define fs first
#define sd second
#define coutY cout<<"YES"<<endl
#define coutN cout<<"NO"<<endl
const int N = 2e5 + 5;
const ll mod = 1e9 + 7;
const ll inf = 0x3f3f3f3f;
int t, n, l, r, s;
ll a[N];
set<ll> vis;
int main()
{
	cin.tie(0); std::ios::sync_with_stdio(false);
	//freopen("c:\\users\\雷神\\desktop\\题\\data.in", "r", stdin);
	//freopen("c:\\users\\雷神\\desktop\\题\\myout.out", "w", stdout);
	cin >> t;
	while (t--)
	{
		bool flag = 0;
		mem(a);
		vis.clear();
		cin >> n >> l >> r >> s;
		int cnt = r - l + 1;
		rep(i, 1, n)
			vis.insert(i);
		if (s < cnt * (1 + cnt) / 2 || s>(2 * n - cnt + 1)* cnt / 2)
		{
			cout << -1 << endl;
			continue;
		}
		else
		{
			int cha = s - cnt * (1 + cnt) / 2;
			int bei = cha / cnt;
			int mo = cha % cnt;
			for (int i = r; i >= l; i--)
			{
				a[i] += i - l + 1 + bei;
				if (mo)
					a[i]++,mo--;
				vis.erase(a[i]);
			}
			rep(i, 1, n)
				if (!a[i])
					a[i] = *vis.begin(), vis.erase(vis.begin());
		}
		rep(i, 1, n)
			cout << a[i] << " ";
		cout << endl;
	}
	return 0;
}



### Prompt

In Cpp, your task is to solve the following problem:

A permutation is a sequence of n integers from 1 to n, in which all the numbers occur exactly once. For example, [1], [3, 5, 2, 1, 4], [1, 3, 2] are permutations, and [2, 3, 2], [4, 3, 1], [0] are not.

Polycarp was given four integers n, l, r (1 ≤ l ≤ r ≤ n) and s (1 ≤ s ≤ (n (n+1))/(2)) and asked to find a permutation p of numbers from 1 to n that satisfies the following condition: 

  * s = p_l + p_{l+1} + … + p_r. 



For example, for n=5, l=3, r=5, and s=8, the following permutations are suitable (not all options are listed): 

  * p = [3, 4, 5, 2, 1]; 
  * p = [5, 2, 4, 3, 1]; 
  * p = [5, 2, 1, 3, 4]. 

But, for example, there is no permutation suitable for the condition above for n=4, l=1, r=1, and s=5.

Help Polycarp, for the given n, l, r, and s, find a permutation of numbers from 1 to n that fits the condition above. If there are several suitable permutations, print any of them.

Input

The first line contains a single integer t (1 ≤ t ≤ 500). Then t test cases follow.

Each test case consist of one line with four integers n (1 ≤ n ≤ 500), l (1 ≤ l ≤ n), r (l ≤ r ≤ n), s (1 ≤ s ≤ (n (n+1))/(2)).

It is guaranteed that the sum of n for all input data sets does not exceed 500.

Output

For each test case, output on a separate line: 

  * n integers — a permutation of length n that fits the condition above if such a permutation exists; 
  * -1, otherwise. 



If there are several suitable permutations, print any of them.

Example

Input


5
5 2 3 5
5 3 4 1
3 1 2 4
2 2 2 2
2 1 1 3


Output


1 2 3 4 5 
-1
1 3 2 
1 2 
-1

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
typedef  long long ll;
#define fat(x) for(auto it:x)
#define  pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define mem(x) memset(x,0,sizeof(x))
#define fs first
#define sd second
#define coutY cout<<"YES"<<endl
#define coutN cout<<"NO"<<endl
const int N = 2e5 + 5;
const ll mod = 1e9 + 7;
const ll inf = 0x3f3f3f3f;
int t, n, l, r, s;
ll a[N];
set<ll> vis;
int main()
{
	cin.tie(0); std::ios::sync_with_stdio(false);
	//freopen("c:\\users\\雷神\\desktop\\题\\data.in", "r", stdin);
	//freopen("c:\\users\\雷神\\desktop\\题\\myout.out", "w", stdout);
	cin >> t;
	while (t--)
	{
		bool flag = 0;
		mem(a);
		vis.clear();
		cin >> n >> l >> r >> s;
		int cnt = r - l + 1;
		rep(i, 1, n)
			vis.insert(i);
		if (s < cnt * (1 + cnt) / 2 || s>(2 * n - cnt + 1)* cnt / 2)
		{
			cout << -1 << endl;
			continue;
		}
		else
		{
			int cha = s - cnt * (1 + cnt) / 2;
			int bei = cha / cnt;
			int mo = cha % cnt;
			for (int i = r; i >= l; i--)
			{
				a[i] += i - l + 1 + bei;
				if (mo)
					a[i]++,mo--;
				vis.erase(a[i]);
			}
			rep(i, 1, n)
				if (!a[i])
					a[i] = *vis.begin(), vis.erase(vis.begin());
		}
		rep(i, 1, n)
			cout << a[i] << " ";
		cout << endl;
	}
	return 0;
}



```