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: import os
  import sys
  from io import BytesIO, IOBase
  #import sys
  #input = sys.stdin.buffer.readline
  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")
  
  
  sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  input = lambda: sys.stdin.readline().rstrip("\r\n")
  #######################################
  from collections import *
  from collections import deque
  from operator import itemgetter , attrgetter
  from decimal import *
  import bisect
  import math
  import heapq as hq
  #import sympy
  MOD=10**9 +7
  
  
  def is_prime(n):
    if n == 2 or n == 3: return True
    if n < 2 or n%2 == 0: return False
    if n < 9: return True
    if n%3 == 0: return False
    r = int(n**0.5)
    # since all primes > 3 are of the form 6n ± 1
    # start with f=5 (which is prime)
    # and test f, f+2 for being prime
    # then loop by 6.
    f = 5
    while f <= r:
  
      if n % f == 0: return False
      if n % (f+2) == 0: return False
      f += 6
    return True
  
  def pow(a,b,m):
      ans=1
      while b:
          if b&1:
              ans=(ans*a)%m
          b//=2
          a=(a*a)%m
      return ans
  #vis=[]
  #graph=[]
  
  def ispalindrome(s):
      if s[:]==s[::-1]:
          return 1
      return 0
  dp=[]
  limit=[]
  v=[]
  def dpdfs(u,t=-1):
      dp[0][u]=0
      dp[1][u]=0
      for i in v[u]:
          if i==t:
  
              continue
          if dp[1][i]==-1:
              dpdfs(i,u)
          dp[0][u]+=max(abs(limit[0][u]-limit[1][i])+dp[1][i],abs(limit[0][u]-limit[0][i])+dp[0][i])
          dp[1][u] += max(abs(limit[1][u] - limit[1][i]) + dp[1][i], abs(limit[1][u] - limit[0][i]) + dp[0][i])
  vis=[]
  f=0
  def dfs(i):
      vis[i]=1
      act[i]=1
      for j in v[i]:
          if act[j]:
              f=1
              #print(-1)
              return -1
          if vis[j]==0:
             if dfs(j)==-1:
                 return -1
      act[i]=0
      ans.append(i)
      return 0
  from queue import PriorityQueue
  
  def z_algorithm(s):
      res = [0] * len(s)
      res[0] = len(s)
      i, j = 1, 0
      while i < len(s):
          while i + j < len(s) and s[j] == s[i + j]:
              j += 1
          res[i] = j
          if j == 0:
              i += 1
              continue
          k = 1
          while i + k < len(s) and k + res[k] < j:
              res[i + k] = res[k]
              k += 1
          i, j = i + k, j - k
      return res
  def gcd(a, b):
      if a == 0:
          return b
      return gcd(b % a, a)
  
  
  # Function to return LCM of two numbers
  def lcm(a, b):
      return (a / gcd(a, b)) * b
  
  
  def gcdExtended(a, b):
      # Base Case
      if a == 0:
          return b, 0, 1
  
      gcd, x1, y1 = gcdExtended(b % a, a)
  
      # Update x and y using results of recursive
      # call
      x = y1 - (b // a) * x1
      y = x1
  
      return gcd, x, y
  import re
  for _ in range(int(input())):
      n=int(input())
      l=[]
      for i in range(2):
          l.append(list(map(int,input().split())))
      ans=math.inf
      l[0][0]=0
      l[1][n-1]=0
      l[1].append(0)
      for i in range(1,n):
          l[0][i]+=l[0][i-1]
          l[1][i]+=l[1][i-1]
      for i in range(n):
          ans=min(ans,max(l[0][n-1]-l[0][i],l[1][i-1]))
      print(ans)