human
Can you provide a Python program to print a single integer — the score of the game if both players play optimally given the following:
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.

Using python3 can you solve the prior task?
assistant
#Here is Python program to print a single integer — the score of the game if both players play optimally.

#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));...