Polycarp has come up with a new game to play with you. He calls it "A missing bigram".

A bigram of a word is a sequence of two adjacent letters in it.

For example, word "abbaaba" contains bigrams "ab", "bb", "ba", "aa", "ab" and "ba".

The game goes as follows. First, Polycarp comes up with a word, consisting only of lowercase letters 'a' and 'b'. Then, he writes down all its bigrams on a whiteboard in the same order as they appear in the word. After that, he wipes one of them off the whiteboard.

Finally, Polycarp invites you to guess what the word that he has come up with was.

Your goal is to find any word such that it's possible to write down all its bigrams and remove one of them, so that the resulting sequence of bigrams is the same as the one Polycarp ended up with.

The tests are generated in such a way that the answer exists. If there are multiple answers, you can print any of them.

Input

The first line contains a single integer t (1 ≤ t ≤ 2000) — the number of testcases.

The first line of each testcase contains a single integer n (3 ≤ n ≤ 100) — the length of the word Polycarp has come up with.

The second line of each testcase contains n-2 bigrams of that word, separated by a single space. Each bigram consists of two letters, each of them is either 'a' or 'b'.

Additional constraint on the input: there exists at least one string such that it is possible to write down all its bigrams, except one, so that the resulting sequence is the same as the sequence in the input. In other words, the answer exists.

Output

For each testcase print a word, consisting of n letters, each of them should be either 'a' or 'b'. It should be possible to write down all its bigrams and remove one of them, so that the resulting sequence of bigrams is the same as the one Polycarp ended up with.

The tests are generated in such a way that the answer exists. If there are multiple answers, you can print any of them. 

Example

Input


4
7
ab bb ba aa ba
7
ab ba aa ab ba
3
aa
5
bb ab bb


Output


abbaaba
abaabaa
baa
bbabb

Note

The first two testcases from the example are produced from the word "abbaaba". As listed in the statement, it contains bigrams "ab", "bb", "ba", "aa", "ab" and "ba".

In the first testcase, the 5-th bigram is removed. 

In the second testcase, the 2-nd bigram is removed. However, that sequence could also have been produced from the word "abaabaa". It contains bigrams "ab", "ba", "aa", "ab", "ba" and "aa". The missing bigram is the 6-th one.

In the third testcase, all of "baa", "aab" and "aaa" are valid answers.
4
7
ab bb ba aa ba
7
ab ba aa ab ba
3
aa
5
bb ab bb

abbaaba
abaabaa
aaa
bbabb

//   package faltu;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
//import java.security.KeyStore.Entry;
import java.util.*;
import java.util.Map.Entry;

public class Main {
    static long LowerBound(long[] a2, long x) { // x is the target value or key
	  int l=-1,r=a2.length;
	  while(l+1<r) {
	    int m=(l+r)>>>1;
	    if(a2[m]>=x) r=m;
	    else l=m;
	  }
	  return r;
	}
    static int UpperBound(int a[], int x) {// x is the key or target value
	    int l=-1,r=a.length;
	    while(l+1<r) {
	       int m=(l+r)>>>1;
	       if(a[m]<=x) l=m;
	       else r=m;
	    }
	    return l+1;
    }
    public static long getClosest(long val1, long val2,long target)
	{
		if (target - val1 >= val2 - target)
		return val2;       
		else
		return val1;       
	}
    static void ruffleSort(long[] a) {
		int n=a.length;
		Random r=new Random();
		for (int i=0; i<a.length; i++) {
			long oi=r.nextInt(n), temp=a[i];
			a[i]=a[(int)oi];
			a[(int)oi]=temp;
		}
		Arrays.sort(a);
	}
    static void ruffleSort(int[] a) {
		int n=a.length;
		Random r=new Random();
		for (int i=0; i<a.length; i++) {
			int oi=r.nextInt(n), temp=a[i];
			a[i]=a[oi];
			a[oi]=temp;
		}
		Arrays.sort(a);
	}
    private int ceilIndex(int input[], int T[], int end, int s){
        int start = 0;
        int middle;
        int len = end;
        while(start <= end){
            middle = (start + end)/2;
            if(middle < len && input[T[middle]] < s && s <= input[T[middle+1]]){
                return middle+1;
            }else if(input[T[middle]] < s){
                start = middle+1;
            }else{
                end = middle-1;
            }
        }
        return -1;
    }
    public static int findIndex(long arr[], long t) 
    { 
        if (arr == null) { 
            return -1; 
        } 
        int len = arr.length; 
        int i = 0; 
        while (i < len) {
            if (arr[i] == t) { 
                return i; 
            } 
            else { 
                i = i + 1; 
            } 
        } 
        return -1; 
    }
    static long gcd(long a, long b)
    {
        if (a == 0)
            return b; 
        return gcd(b % a, a); 
    }
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b; 
        return gcd(b % a, a); 
    }
    static long lcm(long a, long b)
    {
        return (a / gcd(a, b)) * b;
    }
 
    public static int[] swap(int a[], int left, int right)
    {
        int temp = a[left];
        a[left] = a[right];
        a[right] = temp;
        return a;
    }
    	public static void swap(int max0,int max1)
    {
    	int temp=max0;
    	max0=max1;
    	max1=temp;
    }
 
    	public static int[] reverse(int a[], int left, int right)
    {
 
        // Reverse the sub-array
        while (left < right) {
            int temp = a[left];
            a[left++] = a[right];
            a[right--] = temp;
        }
        return a;
    }
    	static class FastReader {
        BufferedReader br;
        StringTokenizer st;
 
        public FastReader()
        {
            br = new BufferedReader(
                new InputStreamReader(System.in));
        }
 
        String next()
        {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
 
        int nextInt() { return Integer.parseInt(next()); }
 
        long nextLong() { return Long.parseLong(next()); }
 
        double nextDouble()
        {
            return Double.parseDouble(next());
        }
 
        String nextLine()
        {
            String str = "";
            try {
                str = br.readLine();
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

//    *******----segement tree implement---*****
//    -------------START--------------------------
    	void buildTree (int[] arr,int[] tree,int start,int end,int treeNode)
    {
    	if(start==end)
    	{
    		tree[treeNode]=arr[start];
    		return;
    	}
    	buildTree(arr,tree,start,end,2*treeNode);
    	buildTree(arr,tree,start,end,2*treeNode+1);
    	tree[treeNode]=tree[treeNode*2]+tree[2*treeNode+1];
    }
    	void updateTree(int[] arr,int[] tree,int start,int end,int treeNode,int idx,int value)
    {
    	if(start==end)
    	{
    		arr[idx]=value;
    		tree[treeNode]=value;
    		return;
    	}
    	int mid=(start+end)/2;
    	if(idx>mid)
    	{
    		updateTree(arr,tree,mid+1,end,2*treeNode+1,idx,value);
    	}
    	else
    	{
    		updateTree(arr,tree,start,mid,2*treeNode,idx,value);
    	}
    	tree[treeNode]=tree[2*treeNode]+tree[2*treeNode+1];
    }
    	static int[][] dp;
//    	disjoint set implementation --start
        static void makeSet(int n)
        {
        	parent=new int[n];
        	rank=new int[n];
        	for(int i=0;i<n;i++)
        	{
        		parent[i]=i;
        		rank[i]=0;
        	}
        }
        static void union(int u,int v)
        {
        	u=findpar(u);
        	v=findpar(v);
        	if(rank[u]<rank[v])parent[u]=v;
        	else if(rank[v]<rank[u])parent[v]=u;
        	else 
        	{
        		parent[v]=u;
        		rank[u]++;
        	}
        }
        
    	private static int findpar(int node)
    	{	
    		if(node==parent[node])return node;
    		return parent[node]=findpar(parent[node]);
    	}
    	static int parent[];
        static int rank[];
//    	*************end
        static Long MOD=(long) (1e9+7);
        private static void solver(String[] a,int n)
        {
        	String ans=a[0];
        	int flag = 1;
    	    for(int i=1;i<n-2;i++) {
    	        if(ans.charAt(ans.length()-1)!=a[i].charAt(0)) {
    	            ans+=a[i];
    	            flag = 0;
    	        }
    	        else
    	        ans+=a[i].charAt(1);
    	    }
        	if(ans.length()!=n)
        	{
        		if(ans.charAt(0)=='a')
        		{
        			while(ans.length()<n)ans+='b';
        		}
        		else 
        		{
        			while(ans.length()<n)ans+='a';
        		}
        	}
        	System.out.println(ans);
        }
        public static void main(String[] args)
        {
        	FastReader s = new FastReader();
	        dp=new int[1001][1001];
	        long tt = s.nextLong();
	        while (tt-->0)
	        {
	        	int n=s.nextInt();
	        	String[] a=new String[n-2];
	        	for(int i=0;i<a.length;i++)
	        	{
	        		a[i]=s.next();
	        	}
	        	solver(a,n);
	        }
        }
        static long power1(long a,long b) {
    	    if(b == 0){
    	        return 1;
    	    }
    	    long ans = power(a,b/2);
    	    ans *= ans;
    	    if(b % 2!=0){
    	        ans *= a;
    	    }
    	    return ans;
    	}
        private static int lps(int m ,int n,String s1,String s2,int[][]mat)
        {
            for(int i=1;i<=m;i++)
            {
                for(int j=1;j<=n;j++)
                {
                    if(s1.charAt(i-1)==s2.charAt(j-1))mat[i][j]=1+mat[i-1][j-1];
                    else mat[i][j]=Math.max(mat[i-1][j],mat[i][j-1]);
                }
            }
            return mat[m][n];
        }
        static String lcs(String X, String Y, int m, int n)
        {
            int[][] L = new int[m+1][n+1];
       
            // Following steps build L[m+1][n+1] in bottom up fashion. Note
            // that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] 
            for (int i=0; i<=m; i++)
            {
                for (int j=0; j<=n; j++)
                {
                    if (i == 0 || j == 0)
                        L[i][j] = 0;
                    else if (X.charAt(i-1) == Y.charAt(j-1))
                        L[i][j] = L[i-1][j-1] + 1;
                    else
                        L[i][j] = Math.max(L[i-1][j], L[i][j-1]);
                }
            }
       
            // Following code is used to print LCS
            int index = L[m][n];
            int temp = index;
       
            // Create a character array to store the lcs string
            char[] lcs = new char[index+1];
            lcs[index] = '\u0000'; // Set the terminating character
       
            // Start from the right-most-bottom-most corner and
            // one by one store characters in lcs[]
            int i = m;
            int j = n;
            while (i > 0 && j > 0)
            {
                // If current character in X[] and Y are same, then
                // current character is part of LCS
                if (X.charAt(i-1) == Y.charAt(j-1))
                {
                    // Put current character in result
                    lcs[index-1] = X.charAt(i-1); 
                      
                    // reduce values of i, j and index
                    i--; 
                    j--; 
                    index--;     
                }
       
                // If not same, then find the larger of two and
                // go in the direction of larger value
                else if (L[i-1][j] > L[i][j-1])
                    i--;
                else
                    j--;
            }
            return String.valueOf(lcs);
            // Print the lcs
//            System.out.print("LCS of "+X+" and "+Y+" is ");
//            for(int k=0;k<=temp;k++)
//                System.out.print(lcs[k]);
        }
        static long lis(long[] aa2, int n)
        {
            long lis[] = new long[n];
            int i, j;
            long max = 0;
            for (i = 0; i < n; i++)
                lis[i] = 1;
            for (i = 1; i < n; i++)
                for (j = 0; j < i; j++)
                    if (aa2[i] >= aa2[j] && lis[i] <= lis[j] + 1)
                        lis[i] = lis[j] + 1;
            for (i = 0; i < n; i++)
                if (max < lis[i])
                    max = lis[i];
     
            return max;
        }
        static boolean isPalindrome(String str)
        {
            int i = 0, j = str.length() - 1;
            while (i < j) {
                if (str.charAt(i) != str.charAt(j))
                    return false;
                i++;
                j--;
            }
            return true;
        }
        static boolean issafe(int i, int j, int r,int c, char ch)
        {
			if (i < 0 || j < 0 || i >= r || j >= c|| ch!= '1')return false;
			else return true;
        }
        static long power(long a, long b)
        {
        	   a %=MOD;
        	    long out = 1;
        	    while (b > 0) {
        	        if((b&1)!=0)out = out * a % MOD;
        	        a = a * a % MOD;
        	        b >>= 1;
        			a*=a;
        	    }
        	    return out;
        }
        static long[] sieve;
   	 	public static void sieve()
   	    {
   	        int nnn=(int) 1e6+1;
   		    	long nn=(int) 1e6;
   		    	sieve=new long[(int) nnn];
   		    	int[] freq=new int[(int) nnn];
   		    	sieve[0]=0;
   		    	sieve[1]=1;
   		    	for(int i=2;i<=nn;i++)
   		    	{
   		    		sieve[i]=1;
   		    		freq[i]=1;
   		    	}
   		    	for(int i=2;i*i<=nn;i++)
   		    	{
   		    		if(sieve[i]==1)
   		    		{
   			    		for(int j=i*i;j<=nn;j+=i)
   			    		{
   			    			if(sieve[j]==1)
   			    			{
   			    				sieve[j]=0;
   			    			}
   			    		}
   		    		}
   		    	}
   	    }

}
class pair{
	long x,y;
	public pair(long x,long y)
	{
		this.x=x;
		this.y=y;
	}
}



