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".
import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.util.AbstractMap;
import java.util.Comparator;
import java.util.HashMap;
import java.util.Map;
import java.util.Map.Entry;
import java.util.PriorityQueue;
import java.util.*;
import java.io.*;

// Java Template Pramod Hosahalli...
@SuppressWarnings("unused")
public class Solution {

	//static FastReader in = new FastReader();
	static BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
	static int mod = 1000000007;

	private static void go() throws Exception{
		int n = pi(in.readLine());
		int[] f = new int[5];
		int sum = 0;
		int[][] fc = new int[n][6];
		for(int i = 0; i < n; i++) { 
			for(char c : in.readLine().toCharArray()) {
				f[c-'a']++;
				fc[i][c-'a']++;
				fc[i][5]++;
				sum++;
			} 
		} 

		int count = n;
		while(count > 0){

			int index = 0;
			for(int i = 1; i < f.length; i++)
				if(f[i] > f[index]) index = i;

			if(f[index] > sum - f[index]){
				break;
			}
			//identify the string which has less number of (index+'a') and more other characters
			int neededIndex = 0;
			for(int i = 1; i < n; i++){
				if(fc[i][5] - fc[i][index] > fc[neededIndex][5] - fc[neededIndex][index])
				neededIndex = i;
			}

			for(int i = 0; i < 5; i++)
			f[i] -= fc[neededIndex][i];
			
			sum = sum - fc[neededIndex][5];
			count--;

			Arrays.fill(fc[neededIndex], Integer.MIN_VALUE);
			fc[neededIndex][index] = 0;
		}

		int index = 0;
		for(int i = 1; i < f.length; i++)
			if(f[i] > f[index]) index = i;

		if(f[index] > 0 && f[index] > sum - f[index])
			print(1,count+"");
		else print(1,"0");
    }

	public static void main(String[] args) throws Exception {
		//int t = in.ni();
        int t = pi(in.readLine());
		for(int _i = 1; _i <= t; _i++){
			go();
		}
		in.close();
	}

	static long mul(long a, long b) {
		return ((a % mod) * (b % mod)) % mod;
	}

	static class FastReader {
		final private int BUFFER_SIZE = 1 << 16;
		private DataInputStream din;
		private byte[] buffer;
		private int bufferPointer, bytesRead;

		public FastReader() {
			din = new DataInputStream(System.in);
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}

		public FastReader(String file_name) throws IOException {
			din = new DataInputStream(new FileInputStream(file_name));
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}

		public String readLine() throws IOException {
			byte[] buf = new byte[64]; // line length
			int cnt = 0, c;
			while ((c = read()) != -1) {
				if (c == '\n') {
					if (cnt != 0) {
						break;
					} else {
						continue;
					}
				}
				buf[cnt++] = (byte) c;
			}
			return new String(buf, 0, cnt);
		}

		public int ni() throws IOException {
			int ret = 0;
			byte c = read();
			while (c <= ' ') {
				c = read();
			}
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');

			if (neg)
				return -ret;
			return ret;
		}

		public long nl() throws IOException {
			long ret = 0;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');
			if (neg)
				return -ret;
			return ret;
		}

		public double nd() throws IOException {
			double ret = 0, div = 1;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();

			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');

			if (c == '.') {
				while ((c = read()) >= '0' && c <= '9') {
					ret += (c - '0') / (div *= 10);
				}
			}

			if (neg)
				return -ret;
			return ret;
		}

		private void fillBuffer() throws IOException {
			bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
			if (bytesRead == -1)
				buffer[0] = -1;
		}

		private byte read() throws IOException {
			if (bufferPointer == bytesRead)
				fillBuffer();
			return buffer[bufferPointer++];
		}

		public void close() throws IOException {
			if (din == null)
				return;
			din.close();
		}
	}

	static String rl() throws Exception{
		return in.readLine();
	}

	static int pi(String s) {
		return Integer.parseInt(s);
	}

	static long pl(String s) {
		return Long.parseLong(s);
	}

	static void print(int newline, String s) {
		System.out.print(s);
		if (newline != 0)
			System.out.println();
		// System.out.flush();
	}

	static void cprint(int x) {
		print(0, "Case #" + x + ": ");
	}
}