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.
import java.io.*;
import java.util.*;

public class Solution {
    //static long[] diff;
    static boolean debug = false;
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        int T = Integer.parseInt(in.readLine());
        HashMap<Long, Integer> hs = new HashMap<>();
        while (T-- > 0) {
            boolean poss = true;
            int N = Integer.parseInt(in.readLine());
            String[] str = (in.readLine()).split(" ");
            //diff = new long[2*N];
            TreeSet<Long> ts = new TreeSet<>();
            int curFreq = 0;
            hs = new HashMap<>();
            for (int i=0; i<2*N; i++) {
                long cur = Long.parseLong(str[i]);
                if (hs.containsKey(cur)) {
                    curFreq = hs.get(cur);
                } else {
                    curFreq = 0;
                }
                hs.put(cur, curFreq+1);
                ts.add(-cur);
            }
            
            Iterator it = hs.entrySet().iterator();
            
            if (debug)
                System.out.println("[");
            while (it.hasNext()) {
                Map.Entry mapElement = (Map.Entry)(it.next());
                long curEle = (long)(mapElement.getKey());
                curFreq = (int)(mapElement.getValue());
                if (debug)
                    System.out.println("{curEle= "+curEle+" curFreq= "+curFreq+"}, ");
                if (curFreq!=2) {
                    poss = false;
                    break;
                }
                
                /*
                if (!hs.containsKey(-curEle) || hs.get(curEle)!=1) {
                    poss = false;
                    break;
                }
                */
            }
            if (debug)
                System.out.println("]");
            
            if (!poss) {
                System.out.println("NO");
                continue;
            }
            
            long oldEle = -1;
            int multiplier = 2*N;
            Iterator<Long> treeIt = ts.iterator();
            int ind = 0;
            long[] xDiff = new long[N-1];
            while (treeIt.hasNext()) {
                long curEle = treeIt.next();
                if (ind == 0) {
                    oldEle = curEle;                    
                    ind++;
                    continue;
                }
                long curDiff = curEle - oldEle; // (-86) - (-104) = 18
                if (debug)
                    System.out.println("curDiff = " + curDiff);
                multiplier -= 2;
                if (curDiff % multiplier != 0) {
                    poss = false;
                    break;
                } else {
                    xDiff[ind-1] = curDiff/multiplier;
                }
        
                oldEle = curEle;                    
                ind++;
            }
            
            if (!poss) {
                System.out.println("NO");
                continue;
            }
            
            //by this point, if poss is true, then ind should = N
            //and oldEle and curEle should both equal the negative of the smallest
            //inputted difference element in input array d
            
            //we need to see if we can choose a value for the smallest positive (or zero)
            //element of array a, such that this determines all the greater elements of
            //array a (through the difference array xDiff), and more importantly
            //such that the corresponding calculated d value for the smallest element of
            //array a is equal to the inputted smallest d value.
            
            int coeff = 2*N;
            long minValue = -1;
            for (int i=N-2; i>-1; i--) {
                oldEle += 2*(i+1)*xDiff[i];
            }
            oldEle = -oldEle;
            if (debug)
                System.out.println("oldEle = " + oldEle);
            if (oldEle <= 0) {
                poss = false;
                //could possibly change it to also include <= 0
            } else if (oldEle % coeff != 0) {
                poss = false;
            } else {
                minValue = oldEle / coeff;
                //minValue should be a non-negative integer
                poss = true;
                System.out.println("YES");
                continue;
            }
            
            if (!poss) {
                System.out.println("NO");
                continue;
            }
            
            
            //if we calculate the difference d for the smallest element (the one that should be in the inputted array d[]), then this difference d = coeff*w + off
            
        }
    }
}