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.*;
import java.util.*;

public class Codeforce {

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

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

        public Reader(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 nextInt() 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 nextLong() 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 nextDouble() 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 int mod = (int) (1e9 + 7);

    public static class pair implements Comparator<pair> {
        int x;
        int y;

        public pair() {

        }

        public pair(int x, int y) {
            this.x = x;
            this.y = y;

        }

        @Override
        public int compare(pair o1, pair o2) {
            return o1.y - o2.y;
        }
    }

    public static long modularpow(long a, long b) {
        long res = 1;
        if (b == 0)
            return res;
        else {
            while (b > 0) {
                if (b % 2 == 1) {
                    res *= a;
                    res %= mod;

                }
                a = a * a;
                a %= mod;
                b /= 2;
            }
            return res % mod;
        }
    }

    public static int gcd(int a, int b) {
        if (b == 0)
            return a;
        else
            return gcd(b, a % b);
    }

    public static String binary(int a) {
        String s1 = "";
        while (a > 0) {
            s1 = a % 2 + s1;
            a /= 2;
        }
        return s1;
    }

    public static int Lower_Bound(long a[], int l, int r, long k) {


        while (r - l > 1) {
            int mid = l + (r - l) / 2;
            if (a[mid] <= k) l = mid;
            else
                r = mid;

        }
        return l;
    }

    public static int Upper_Bound(int a[], int l, int r, int k) {

        while (r - l > 1) {
            int mid = (l + r) / 2;
            if (a[mid] <= k)
                l = mid;
            else
                r = mid;
        }
        return l + 1;
    }
public static  int fun(int ch,int a[][],int tot[],int n){
        ArrayList<Integer>ar=new ArrayList<>();
        for(int i=0;i<n;i++){
            ar.add((a[i][ch]-(tot[i]-a[i][ch])));
        }
        Collections.sort(ar,Collections.reverseOrder());
        int cou=0;
        int ans=0;
        for(int i:ar){
            cou+=i;
            if(cou>0){
                ans++;
            }
            else
                break;
        }
return ans;
}
    public static void main(String arg[]) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(br.readLine());
        StringBuilder sb = new StringBuilder();
        while (t-- > 0) {
               int n=Integer.parseInt(br.readLine());
               int a[][]=new int[n][5];
               int tot[]=new int[n];
               for(int i=0;i<n;i++){
                   String s=br.readLine();
                   tot[i]+=s.length();
                   for(int j=0;j<s.length();j++){
                       a[i][s.charAt(j)-'a']++;
                   }
               }
               int ans=0;
               for(int i=0;i<5;i++){
                   ans=Math.max(ans,fun(i,a,tot,n));
               }
               sb.append(ans+"\n");
                       }
        System.out.println(sb);
    }
}