description: 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. 
solution: """
  // Author : snape_here - Susanta Mukherjee
       
   """
  
  from __future__ import division, print_function
   
  import os,sys
  from io import BytesIO, IOBase
   
  if sys.version_info[0] < 3:
      from __builtin__ import xrange as range
      from future_builtins import ascii, filter, hex, map, oct, zip
   
  def ii(): return int(input())
  def fi(): return float(input())
  def si(): return input()
  def msi(): return map(str,input().split())
  def mi(): return map(int,input().split())
  def li(): return list(mi())
  def lsi(): return list(msi())
   
  def read():
      sys.stdin = open('input.txt', 'r')  
      sys.stdout = open('output.txt', 'w') 
   
  def gcd(x, y):
      while y:
          x, y = y, x % y
      return x
  
  def lcm(x, y):
      return (x*y)//(gcd(x,y))
  
  mod=1000000007
  
  def modInverse(b,m): 
      g = gcd(b, m)  
      if (g != 1):         
          return -1
      else:          
          return pow(b, m - 2, m) 
  
  def ceil2(x,y):
      if x%y==0:
          return x//y
      else:
          return x//y+1
  
  def modu(a,b,m): 
  
      a = a % m 
      inv = modInverse(b,m) 
      if(inv == -1): 
          return -999999999
      else: 
          return (inv*a)%m
  
  from math import log,factorial,cos,tan,sin,radians,floor,sqrt,ceil
  
  import bisect
  import random
  import string 
  
  from decimal import *
  
  getcontext().prec = 50
  
  abc="abcdefghijklmnopqrstuvwxyz"
  
  pi=3.141592653589793238
  
  def gcd1(a):
      if len(a) == 1:
          return a[0]
      ans = a[0]
      for i in range(1,len(a)):
          ans = gcd(ans,a[i])
      return ans
  
  def mykey(x):
      return len(x)
  
  def main():
  
      for _ in range(ii()):
          n=ii()
          l = []
          for i in range(n):
              a=li()
              k=a[0]
              b=a[1:]
              x = b[-1]
              c = [0]*k
              for i in range(k):
                  c[i] = b[i]+1-i
              l.append([max(c),k])
          l.sort()
          #print(l)
          ans = l[0][0]
          c = l[0][1]
          for i in range(1,n):
              ans = max(l[i][0]-c,ans)
              c += l[i][1]
          print(ans)
  
  
  # region fastio
  
  # template taken from https://github.com/cheran-senthil/PyRival/blob/master/templates/template.py
  
  BUFSIZE = 8192
   
  
  class FastIO(IOBase):
      newlines = 0
   
      def __init__(self, file):
          self._fd = file.fileno()
          self.buffer = BytesIO()
          self.writable = "x" in file.mode or "r" not in file.mode
          self.write = self.buffer.write if self.writable else None
   
      def read(self):
          while True:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              if not b:
                  break
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines = 0
          return self.buffer.read()
   
      def readline(self):
          while self.newlines == 0:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              self.newlines = b.count(b"\n") + (not b)
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines -= 1
          return self.buffer.readline()
   
      def flush(self):
          if self.writable:
              os.write(self._fd, self.buffer.getvalue())
              self.buffer.truncate(0), self.buffer.seek(0)
   
   
  class IOWrapper(IOBase):
      def __init__(self, file):
          self.buffer = FastIO(file)
          self.flush = self.buffer.flush
          self.writable = self.buffer.writable
          self.write = lambda s: self.buffer.write(s.encode("ascii"))
          self.read = lambda: self.buffer.read().decode("ascii")
          self.readline = lambda: self.buffer.readline().decode("ascii")
   
   
  def print(*args, **kwargs):
      """Prints the values to a stream, or to sys.stdout by default."""
      sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
      at_start = True
      for x in args:
          if not at_start:
              file.write(sep)
          file.write(str(x))
          at_start = False
      file.write(kwargs.pop("end", "\n"))
      if kwargs.pop("flush", False):
          file.flush()
   
   
  if sys.version_info[0] < 3:
      sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
  else:
      sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
   
  input = lambda: sys.stdin.readline().rstrip("\r\n")
   
  # endregion
  
   
  if __name__ == "__main__":
      #read()
      main()