Stephen Queen wants to write a story. He is a very unusual writer, he uses only letters 'a', 'b', 'c', 'd' and 'e'!

To compose a story, Stephen wrote out n words consisting of the first 5 lowercase letters of the Latin alphabet. He wants to select the maximum number of words to make an interesting story.

Let a story be a sequence of words that are not necessarily different. A story is called interesting if there exists a letter which occurs among all words of the story more times than all other letters together.

For example, the story consisting of three words "bac", "aaada", "e" is interesting (the letter 'a' occurs 5 times, all other letters occur 4 times in total). But the story consisting of two words "aba", "abcde" is not (no such letter that it occurs more than all other letters in total).

You are given a sequence of n words consisting of letters 'a', 'b', 'c', 'd' and 'e'. Your task is to choose the maximum number of them to make an interesting story. If there's no way to make a non-empty story, output 0.

Input

The first line contains one integer t (1 ≤ t ≤ 5000) — the number of test cases. Then t test cases follow.

The first line of each test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of the words in the sequence. Then n lines follow, each of them contains a word — a non-empty string consisting of lowercase letters of the Latin alphabet. The words in the sequence may be non-distinct (i. e. duplicates are allowed). Only the letters 'a', 'b', 'c', 'd' and 'e' may occur in the words.

It is guaranteed that the sum of n over all test cases doesn't exceed 2 ⋅ 10^5; the sum of the lengths of all words over all test cases doesn't exceed 4 ⋅ 10^5.

Output

For each test case, output the maximum number of words that compose an interesting story. Print 0 if there's no way to make a non-empty interesting story.

Example

Input


6
3
bac
aaada
e
3
aba
abcde
aba
2
baba
baba
4
ab
ab
c
bc
5
cbdca
d
a
d
e
3
b
c
ca


Output


3
2
0
2
3
2

Note

In the first test case of the example, all 3 words can be used to make an interesting story. The interesting story is "bac aaada e".

In the second test case of the example, the 1-st and the 3-rd words can be used to make an interesting story. The interesting story is "aba aba". Stephen can't use all three words at the same time.

In the third test case of the example, Stephen can't make a non-empty interesting story. So the answer is 0.

In the fourth test case of the example, Stephen can use the 3-rd and the 4-th words to make an interesting story. The interesting story is "c bc".
/*
                       _oo0oo_
                      o8888888o
                      88" . "88
                      (| -_- |)
                      0\  =  /0
                    ___/`---'\___
                  .' \\|     |// '.
                 / \\|||  :  |||// \
                / _||||| -:- |||||- \
               |   | \\\  -  /// |   |
               | \_|  ''\---/''  |_/ |
               \  .-\__  '-'  ___/-. /
             ___'. .'  /--.--\  `. .'___
          ."" '<  `.___\_<|>_/___.' >' "".
         | | :  `- \`.;`\ _ /`;.`/ - ` : | |
         \  \ `_.   \_ __\ /__ _/   .-` /  /
     =====`-.____`.___ \_____/___.-`___.-'=====
                       `=---='
 
*/
import java.util.function.Consumer;
import java.util.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.math.RoundingMode;
import java.text.DecimalFormat;
import java.io.*;
import java.lang.Math.*;
 public class KickStart2020{
	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());}
        float nextFloat() {return Float.parseFloat(next());}
        String nextLine() {
            String str = "";
            try {str = br.readLine();}
            catch (IOException e) { e.printStackTrace();}
            return str; }}
	static boolean isBracketSequence(String s, int a, int b) {
		Stack<Character> ss = new Stack<>();
		boolean hachu = true;
		for(int i = a; i <= b; i++) {
			if(s.charAt(i) == ')' && ss.isEmpty()) {hachu = false; break;}
			if(s.charAt(i) == '(') ss.add('(');
			else ss.pop();
		}
		return ss.empty() && hachu;
	}
	static String reverseOfString(String a) {
		StringBuilder ssd = new StringBuilder();
		for(int i = a.length() - 1; i >= 0; i--) {
			ssd.append(a.charAt(i));
		}
		return ssd.toString();
	}
	static char[] reverseOfChar(char a[]) {
		char b[] = new char[a.length];
		int j = 0;
		for(int i = a.length - 1; i >= 0; i--) {
			b[i] = a[j];
			j++;
		}
		return b;
	}
	static boolean isPalindrome(char a[]) {
		boolean hachu = true;
		for(int i = 0; i <= a.length / 2; i++) {
			if(a[i] != a[a.length - 1 - i]) {
				hachu = false;
				break;
			}
		}
		return hachu;
	} 
	static long gcd(long a, long b)
	{
	    if (b == 0)
	        return a;
	    return gcd(b, a % b);
	     
	}
	static long powermod(long x, long y, long mod){
	    long ans = 1;
	    x = x % mod; 
	    if (x == 0)
	      return 0; 
	    int i = 1;
	    while (y > 0){
	      if ((y & 1) != 0)
	        ans = (ans * x) % mod;
	      y = y >> 1; 
	      x = (x * x) % mod;
	    }
	    return ans;
	  }
	static long power(long x, long y){
	    long ans = 1;
	    if (x == 0)
	      return 0; 
	    int i = 1;
	    while (y > 0){
	      if ((y & 1) != 0)
	        ans = (ans * x);
	      y = y >> 1; 
	      x = (x * x);
	    }
	    return ans;
	  }
	static boolean check(String a) {
		boolean hachu = true;
		for(int i = 0; i < a.length(); i++) {
			if(a.charAt(0) != a.charAt(i)) {hachu = false; break;}
		}
		return hachu;
	}
	 public static class Pair implements Comparable<Pair> {
	        public  int index;
	        public  int value;
	 
	        public Pair(int index, int value) {
	            this.index = index;
	            this.value = value;
	        }
	        @Override
	        public int compareTo(Pair other) {
	            //multiplied to -1 as the author need descending sort order
	            return   -1 * Integer.valueOf(this.value).compareTo(other.value);
	        }
	 }
	 static boolean equalString(int i, int j, int arr[], String b) {
		 int brr[] = new int[26];
		 for(int k = i; k <= j; k++) brr[b.charAt(k) - 'a']++;
		 for(int k = 0; k < 26; k++) {
			 if(arr[k] != brr[k]) return false;
		 }
		 return true;
	 }
	 static boolean cequalArray(String a, String b) {
		  int count[] = new int[26];
		  int count1[] = new int[26];
		  for(int i = 0; i < a.length(); i++) count[a.charAt(i) - 'a']++;
		  for(int i = 0; i < a.length(); i++) count1[b.charAt(i) - 'a']++;
		  for(int i = 0; i < 26; i++) if(count[i] != count1[i]) return false;
		  return true;
	 }
	 static boolean isPrime(long d) {
		 if(d == 1) return true;
		 for(int i = 2; i <= (long)Math.sqrt(d); i++) {
			 if(d % i == 0) return false;
		 }
		 return true;
	 }
	 static int getAns(char a, String arr[]) {
		 int n = arr.length;
		 int brr[] = new int[n];
		 for(int i = 0; i < n; i++) {
		 for(char e: arr[i].toCharArray()) {
			 if(e == a) brr[i]++;
			 else brr[i]--;
		 }}
		 Arrays.sort(brr);
		 int sum = 0;
		 int i = n - 1;
		 int count = 0;
		 while(i >= 0 && brr[i] + sum > 0) {
			 sum += brr[i];
			 count++;
			 i--;
		 }
		 return count;
	 }
		public static void main(String[] args) throws Exception{
			FastReader sc = new FastReader();
			PrintWriter out = new PrintWriter(System.out);
              int t = sc.nextInt();
              outerloop:
           while(t-- > 0) {
                int n = sc.nextInt();
               String arr[] = new String[n];
               for(int i = 0; i < n; i++) 
            	   arr[i] = sc.next();
               int ans = -1;
               for(char i = 'a'; i <= 'e'; i++) {
            	   int z = getAns(i, arr);
            	   ans = Math.max(ans, z);
               }
               out.println(ans);
                 }
                 out.close();
			}
		}