description: Alice and Bob are playing a game on a matrix, consisting of 2 rows and m columns. The cell in the i-th row in the j-th column contains a_{i, j} coins in it.
  
  Initially, both Alice and Bob are standing in a cell (1, 1). They are going to perform a sequence of moves to reach a cell (2, m).
  
  The possible moves are: 
  
    * Move right — from some cell (x, y) to (x, y + 1); 
    * Move down — from some cell (x, y) to (x + 1, y). 
  
  
  
  First, Alice makes all her moves until she reaches (2, m). She collects the coins in all cells she visit (including the starting cell).
  
  When Alice finishes, Bob starts his journey. He also performs the moves to reach (2, m) and collects the coins in all cells that he visited, but Alice didn't.
  
  The score of the game is the total number of coins Bob collects.
  
  Alice wants to minimize the score. Bob wants to maximize the score. What will the score of the game be if both players play optimally?
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of testcases.
  
  Then the descriptions of t testcases follow.
  
  The first line of the testcase contains a single integer m (1 ≤ m ≤ 10^5) — the number of columns of the matrix.
  
  The i-th of the next 2 lines contain m integers a_{i,1}, a_{i,2}, ..., a_{i,m} (1 ≤ a_{i,j} ≤ 10^4) — the number of coins in the cell in the i-th row in the j-th column of the matrix.
  
  The sum of m over all testcases doesn't exceed 10^5.
  
  Output
  
  For each testcase print a single integer — the score of the game if both players play optimally.
  
  Example
  
  Input
  
  
  3
  3
  1 3 7
  3 5 1
  3
  1 3 9
  3 5 1
  1
  4
  7
  
  
  Output
  
  
  7
  8
  0
  
  Note
  
  The paths for the testcases are shown on the following pictures. Alice's path is depicted in red and Bob's path is depicted in blue.
  
  <image>
solution: #Mamma don't raises quitter.................................................
  from collections import deque as de
  import math
  from math import sqrt as sq
  from math import floor as fl
  from math import ceil as ce
  from sys import stdin, stdout
  import re
  from collections import Counter as cnt
  from functools import  reduce
  
  from itertools import groupby as gb
  #from fractions import Fraction as fr
  from bisect import bisect_left as bl, bisect_right as br
  
  def factors(n):    
      return set(reduce(list.__add__, 
                  ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
      
  class My_stack():
      def __init__(self):
          self.data = []
      def my_push(self, x):
          return (self.data.append(x))
      def my_pop(self):
          return (self.data.pop())
      
      def my_peak(self):
          return (self.data[-1])
      def my_contains(self, x):
          return (self.data.count(x))
      def my_show_all(self):
          return (self.data)
      def isEmpty(self):
        return len(self.data)==0
  
  arrStack = My_stack() 
  
  #decimal to binary   
  def decimalToBinary(n): 
      return bin(n).replace("0b", "")
  #binary to decimal
  def binarytodecimal(n):
      return int(n,2)
  
  def isPrime(n) : 
  	if (n <= 1) : 
  		return False
  	if (n <= 3) : 
  		return True
  
  	
  	if (n % 2 == 0 or n % 3 == 0) : 
  		return False
  
  	i = 5
  	while(i * i <= n) : 
  		if (n % i == 0 or n % (i + 2) == 0) : 
  			return False
  		i = i + 6
  
  	return True
  
  def get_prime_factors(number):
      prime_factors = []
      while number % 2 == 0:
          prime_factors.append(2)
          number = number / 2
      for i in range(3, int(math.sqrt(number)) + 1, 2):
          while number % i == 0:
              prime_factors.append(int(i))
              number = number / i
  
      if number > 2:
          prime_factors.append(int(number))
      return prime_factors
  
  def get_frequency(list):
      dic={}
      for ele in list:
          if ele in dic:
              dic[ele] += 1
          else:
              dic[ele] = 1
      return dic
  def Log2(x): 
      return (math.log10(x) / 
              math.log10(2)); 
  
   
  # Function to get product of digits
  def getProduct(n):
   
      product = 1
   
      while (n != 0):
          product = product * (n % 10)
          n = n // 10
   
      return product
  
  
  #function to find LCM of two numbers
  def lcm(x,y):
     lcm = (x*y)//math.gcd(x,y)
     return lcm
  
  def isPowerOfTwo(n): 
      return (math.ceil(Log2(n)) == math.floor(Log2(n))); 
  #to check whether the given sorted sequnce is forming an AP or not....
  def checkisap(list):
      d=list[1]-list[0]
      for i in range(2,len(list)):
          temp=list[i]-list[i-1]
          if temp !=d:
              return False
      return True
          
  #seive of erathanos
  def primes_method5(n):
      out ={}
      sieve = [True] * (n+1)
      for p in range(2, n+1):
          if (sieve[p]):
              out[p]=1
              for i in range(p, n+1, p):
                  sieve[i] = False
      return out
  #function to get the sum of digits
  def getSum(n): 
        
      strr = str(n)
      list_of_number = list(map(int, strr.strip()))
      return sum(list_of_number)
  
  
  #ceil  function gives wrong answer after 10^17 so i have to create my own :)
  # because i don't want to doubt on my solution of 900-1000 problem set.
  def ceildiv(x,y): 
      return (x+y-1)//y 
    
  def di():return map(int, input().split())
  def ii():return int(input())
  def li():return list(map(int, input().split()))
  def si():return list(map(str, input()))
  def indict():
      dic = {}
      for index, value in enumerate(input().split()):
          dic[int(value)] = int(index)+1
      return dic
  def frqdict(): 
      # by default it is for integer input. :)
      dic={}
      for index, value in enumerate(input()):
          if value not in dic:
              dic[value] =1
          else:
              dic[value] +=1
      return dic
  
  #inp = open("input.txt","r")
  #out = open("output.txt","w")
  #Here we go......................
  #practice like your never won
  #perform like you never lost
  t=ii()
  while t:
      t-=1
      n=ii()
      a=li()
      b=li()
      if n==1:
          print(0)
      elif n==2:
          print(min(a[1], b[0]))
      else:
          sumb=sum(b)
          suma=sum(a)
          tempa=suma-a[0]
          tempb=0
          mn=max(tempa, tempb)
          for i in range(1, n):
              tempa-=a[i]
              tempb+=b[i-1]
              tempmax=max(tempa, tempb)
              if tempmax < mn:
                  mn=tempmax
          print(mn)
     
          
          
  
  
  
  
  
  
  
  
  
          
  
          
      
  
  
  
              
  
      
      
      
  
  
  
              
  
              
  
  
      
          
  
  
          
      
  
              
          
      
  
                  
                  
                  
          
          
          
  
              
  
          
  
  
  
  
  
                      
                  
              
          
  
              
      
          
      
      
  
      
          
      
   
  
      
          
  
  
      
          
  
  