Long time ago there was a symmetric array a_1,a_2,…,a_{2n} consisting of 2n distinct integers. Array a_1,a_2,…,a_{2n} is called symmetric if for each integer 1 ≤ i ≤ 2n, there exists an integer 1 ≤ j ≤ 2n such that a_i = -a_j.

For each integer 1 ≤ i ≤ 2n, Nezzar wrote down an integer d_i equal to the sum of absolute differences from a_i to all integers in a, i. e. d_i = ∑_{j = 1}^{2n} {|a_i - a_j|}.

Now a million years has passed and Nezzar can barely remember the array d and totally forget a. Nezzar wonders if there exists any symmetric array a consisting of 2n distinct integers that generates the array d.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. 

The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5).

The second line of each test case contains 2n integers d_1, d_2, …, d_{2n} (0 ≤ d_i ≤ 10^{12}).

It is guaranteed that the sum of n over all test cases does not exceed 10^5.

Output

For each test case, print "YES" in a single line if there exists a possible array a. Otherwise, print "NO".

You can print letters in any case (upper or lower).

Example

Input


6
2
8 12 8 12
2
7 7 9 11
2
7 11 7 11
1
1 1
4
40 56 48 40 80 56 80 48
6
240 154 210 162 174 154 186 240 174 186 162 210


Output


YES
NO
NO
NO
NO
YES

Note

In the first test case, a=[1,-3,-1,3] is one possible symmetric array that generates the array d=[8,12,8,12].

In the second test case, it can be shown that there is no symmetric array consisting of distinct integers that can generate array d.
#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define N 200010
#define ll long long
ll t,n,d[N],a[N],sum,di,lst;
using namespace std;
int main()
{
	cin>>t;
	while(t--)
	{
		cin>>n;
		for(ll i=1;i<=2*n;i++)
		{
			scanf("%lld",&d[i]);
		}
		sort(d+1,d+2*n+1);
		bool flag=true;
		for(ll i=2;i<=2*n;i+=2)
		{
			if(d[i]%2!=0 || d[i]!=d[i-1] || d[i]<=d[i-2])
			{
				//printf("NO\n");
				flag=false;
				break;
			}
		}
		if(!flag)
		{
			printf("NO\n");
			continue;
		}
		//	continue;
		sum=0;
		di=2*n;
		lst=1e18;
		for(ll i=2*n;i>=2;i-=2)
		{
			if((d[i]-2*sum)%di != 0)
			{
				flag=false;
				break;
			}
			a[i] = (d[i]-2*sum)/di;
			if(a[i]>=lst || a[i]<=0)
			{
				flag=false;
				break;
			}
			lst = a[i];
			sum += a[i];
			di -= 2;
		}
		if(!flag)
		{
			printf("NO\n");
			continue;
		}
		else
			printf("YES\n");
	}
	return 0;
}
