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<bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define f(i,x,y) for(int i=x;i<=y;i++)
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//priority_queue<int>Q;
//unordered_map<string,int> Q;
const int MAX=3e5+10;
ll a[MAX];
void solve() {
	ll n;
	cin>>n;
	for(int i=1; i<=n*2; i++) {
		cin>>a[i];
	}
	int flag1=1;
	sort(a+1,a+1+n*2,greater<ll>());
	ll last;
	int v=-1;
	ll flag=0;
	ll cnt=n;
	for(int i=1; i<=n*2; i+=2) {
		if(a[i]!=a[i+1]||a[i]<2||a[i]%2) {
			flag1=0;
			cout<<"NO"<<endl;
			break;
		}
		a[i]-=flag;
		if((v!=-1&&a[i]>=last*cnt)||a[i]<=0) {
			flag1=0;
			cout<<"NO"<<endl;
			break;
		}
		if(a[i]%cnt==0) {
			v=1;
			last=a[i]/cnt;
		} else {
			flag1=0;
			cout<<"NO"<<endl;
			break;
		}
		if(last&1)
		{
			flag1=0;
			cout<<"NO"<<endl;
		    break;
		}
		flag+=last;
		cnt--;
	}
	if(flag1)
		cout<<"YES"<<endl;
}
int main() {
	IOS;
	int t;
	cin>>t;
	while(t--) {
		solve();
	}
}
