In a certain video game, the player controls a hero characterized by a single integer value: power. The hero will have to beat monsters that are also characterized by a single integer value: armor.

On the current level, the hero is facing n caves. To pass the level, the hero must enter all the caves in some order, each cave exactly once, and exit every cave safe and sound. When the hero enters cave i, he will have to fight k_i monsters in a row: first a monster with armor a_{i, 1}, then a monster with armor a_{i, 2} and so on, finally, a monster with armor a_{i, k_i}.

The hero can beat a monster if and only if the hero's power is strictly greater than the monster's armor. If the hero can't beat the monster he's fighting, the game ends and the player loses. Note that once the hero enters a cave, he can't exit it before he fights all the monsters in it, strictly in the given order.

Each time the hero beats a monster, the hero's power increases by 1.

Find the smallest possible power the hero must start the level with to be able to enter all the caves in some order and beat all the monsters.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^5). Description of the test cases follows.

The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of caves.

The i-th of the next n lines contains an integer k_i (1 ≤ k_i ≤ 10^5) — the number of monsters in the i-th cave, followed by k_i integers a_{i, 1}, a_{i, 2}, …, a_{i, k_i} (1 ≤ a_{i, j} ≤ 10^9) — armor levels of the monsters in cave i in order the hero has to fight them.

It is guaranteed that the sum of k_i over all test cases does not exceed 10^5.

Output

For each test case print a single integer — the smallest possible power the hero must start the level with to be able to enter all the caves in some order and beat all the monsters.

Example

Input


2
1
1 42
2
3 10 15 8
2 12 11


Output


43
13

Note

In the first test case, the hero has to beat a single monster with armor 42, it's enough to have power 43 to achieve that.

In the second test case, the hero can pass the level with initial power 13 as follows: 

  * enter cave 2: 
    * beat a monster with armor 12, power increases to 14; 
    * beat a monster with armor 11, power increases to 15; 
  * enter cave 1: 
    * beat a monster with armor 10, power increases to 16; 
    * beat a monster with armor 15, power increases to 17; 
    * beat a monster with armor 8, power increases to 18. 
 //package eround101;

import java.util.*;
import java.io.*;
import java.lang.*;
import java.util.StringTokenizer;

public class C101 {

    static HritikScanner sc = new HritikScanner();
    static PrintWriter pw = new PrintWriter(System.out, true);
    final static int MOD = (int) 1e9 + 7;
    static StringBuilder sb = new StringBuilder();

    public static void main(String[] args) {

        int t = ni();
        while (t-- > 0) {
            solve();
        }
    }

    static void solve() {
        int cave = ni();
        int summons = 0;
        PriorityQueue<int[]> pq = new PriorityQueue<>((a, b)-> b[0] - a[0]);
        for(int i = 0; i < cave; i++)
        {
            int n = ni();
            summons += n;
            int maxEntry = -1;
            for(int j = 0; j < n; j++)
            {
                int armor = ni();
                maxEntry = Math.max(maxEntry, armor+1-j);
            }
            pq.add(new int[]{maxEntry, n});
        }
        int[] ar = pq.poll();
        summons = summons - ar[1];
        pl(ar[0] - summons);
        
    }
    
    
    
    
    /////////////////////////////////////////////////////////////////////////////////
    static int[] nextIntArray(int n) {
        int[] arr = new int[n];
        int i = 0;
        while (i < n) {
            arr[i++] = ni();
        }
        return arr;
    }

    static long[] nextLongArray(int n) {
        long[] arr = new long[n];
        int i = 0;
        while (i < n) {
            arr[i++] = nl();
        }
        return arr;
    }

    static int[] nextIntArray1(int n) {
        int[] arr = new int[n + 1];
        int i = 1;
        while (i <= n) {
            arr[i++] = ni();
        }
        return arr;
    }

    /////////////////////////////////////////////////////////////////////////////////
    static int ni() {
        return sc.nextInt();
    }

    static long nl() {
        return sc.nextLong();
    }

    static double nd() {
        return sc.nextDouble();
    }

    /////////////////////////////////////////////////////////////////////////////////
    static void pl() {
        pw.println();
    }

    static void p(Object o) {
        pw.print(o+ " ");
    }

    static void pl(Object o) {
        pw.println(o);
    }

    static void psb(StringBuilder sb) {
        pw.print(sb);
    }

    static void pa(Object arr[]) {
        for (Object o : arr) {
            p(o);
        }
        pl();
    }

    static void pa(int arr[]) {
        for (int o : arr) {
            p(o);
        }
        pl();
    }

    static void pa(long arr[]) {
        for (long o : arr) {
            p(o);
        }
        pl();
    }

    static void pa(double arr[]) {
        for (double o : arr) {
            p(o);
        }
        pl();
    }

    static void pa(char arr[]) {
        for (char o : arr) {
            p(o);
        }
        pl();
    }

    static void pa(List list) {
        for (Object o : list) {
            p(o);
        }
        pl();
    }

    static void pa(Object[][] arr) {

        for (int i = 0; i < arr.length; ++i) {
            for (Object o : arr[i]) {
                p(o);
            }
            pl();
        }
    }

    static void pa(int[][] arr) {
        for (int i = 0; i < arr.length; ++i) {
            for (int o : arr[i]) {
                p(o);
            }
            pl();
        }
    }

    static void pa(long[][] arr) {
        for (int i = 0; i < arr.length; ++i) {
            for (long o : arr[i]) {
                p(o);
            }
            pl();
        }
    }

    static void pa(char[][] arr) {
        for (int i = 0; i < arr.length; ++i) {
            for (char o : arr[i]) {
                p(o);
            }
            pl();
        }
    }

    static void pa(double[][] arr) {
        for (int i = 0; i < arr.length; ++i) {
            for (double o : arr[i]) {
                p(o);
            }
            pl();
        }
    }

/////////////////////////////////////////////////////////////////////////////////
    static void print(Object s) {
        System.out.println(s);
    }

    /////////////////////////////////////////////////////////////////////////////////
    //-----------HritikScanner class for faster input----------//
    static class HritikScanner {

        BufferedReader br;
        StringTokenizer st;

        public HritikScanner() {
            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;
        }
    }

    //////////////////////////////////////////////////////////////////
    public static class Pair implements Comparable<Pair> {

        int num, row,col;

        Pair(int num, int row, int col) {
            this.num = num;
            this.row = row;
            this.col = col;
        }

        public int get1() {
            return num;
        }

        public int get2() {
            return row;
        }

        public int compareTo(Pair A) {
            return this.num - A.num;
        }
        
        public String toString()
        {
            return num +" "+ row+" "+col;
        }
    }

    //////////////////////////////////////////////////////////////////
    // Function to return gcd of a and b time complexity O(log(a+b))
    static long gcd(long a, long b) {
        if (a == 0) {
            return b;
        }
        return gcd(b % a, a);
    }

    // method to return LCM of two numbers 
    static long lcm(int a, int b) {
        return (a / gcd(a, b)) * b;
    }

    //////////////////////////////////////////////////////////////////
    static boolean isPrime(long n) {
        // Corner cases 
        if (n <= 1) {
            return false;
        }
        if (n <= 3) {
            return true;
        }

        // This is checked so that we can skip  
        // middle five numbers in below loop 
        if (n % 2 == 0 || n % 3 == 0) {
            return false;
        }

        for (long i = 5; i * i <= n; i = i + 6) {
            if (n % i == 0 || n % (i + 2) == 0) {
                return false;
            }
        }

        return true;
    }
    //////////////////////////////////////////////////////////////////

    static boolean isPowerOfTwo(long n) {
        if (n == 0) {
            return false;
        }

        return (long) (Math.ceil((Math.log(n) / Math.log(2))))
                == (long) (Math.floor(((Math.log(n) / Math.log(2)))));
    }

    public static long getFact(int n) {
        long ans = 1;
        while (n > 0) {
            ans *= n;
            ans %= MOD;
            n--;
        }
        return ans;
    }

    public static long pow(long n, int pow) {
        if (pow == 0) {
            return 1;
        }
        long temp = pow(n, pow / 2) % MOD;
        temp *= temp;
        temp %= MOD;
        if (pow % 2 == 1) {
            temp *= n;
        }
        temp %= MOD;
        return temp;

    }

    public static long nCr(int n, int r) {
        long ans = 1;
        int temp = n - r;
        while (n > temp) {
            ans *= n;
            ans %= MOD;
            n--;
        }
        ans *= pow(getFact(r) % MOD, MOD - 2) % MOD;
        ans %= MOD;
        return ans;
    }

    //////////////////////////////////////////////////////////////////
    // method returns Nth power of A 
    static double nthRoot(int A, int N) {

        // intially guessing a random number between 
        // 0 and 9 
        double xPre = Math.random() % 10;

        // smaller eps, denotes more accuracy 
        double eps = 0.001;

        // initializing difference between two 
        // roots by INT_MAX 
        double delX = 2147483647;

        // xK denotes current value of x 
        double xK = 0.0;

        // loop untill we reach desired accuracy 
        while (delX > eps) {
            // calculating current value from previous 
            // value by newton's method 
            xK = ((N - 1.0) * xPre
                    + (double) A / Math.pow(xPre, N - 1)) / (double) N;
            delX = Math.abs(xK - xPre);
            xPre = xK;
        }

        return xK;
    }

}
