During cleaning the coast, Alice found n piles of stones. The i-th pile has a_i stones.

Piles i and i + 1 are neighbouring for all 1 ≤ i ≤ n - 1. If pile i becomes empty, piles i - 1 and i + 1 doesn't become neighbouring.

Alice is too lazy to remove these stones, so she asked you to take this duty. She allowed you to do only the following operation: 

  * Select two neighboring piles and, if both of them are not empty, remove one stone from each of them. 



Alice understands that sometimes it's impossible to remove all stones with the given operation, so she allowed you to use the following superability: 

  * Before the start of cleaning, you can select two neighboring piles and swap them. 



Determine, if it is possible to remove all stones using the superability not more than once.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.

The first line of each test case contains the single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of piles.

The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the number of stones in each pile.

It is guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.

Output

For each test case, print YES or NO — is it possible to remove all stones using the superability not more than once or not.

Example

Input


5
3
1 2 1
3
1 1 2
5
2 2 2 1 3
5
2100 1900 1600 3000 1600
2
2443 2445


Output


YES
YES
YES
YES
NO

Note

In the first test case, you can remove all stones without using a superability: [1, 2, 1] → [1, 1, 0] → [0, 0, 0].

In the second test case, you can apply superability to the second and the third piles and then act like in the first testcase.

In the third test case, you can apply superability to the fourth and the fifth piles, thus getting a = [2, 2, 2, 3, 1].

In the fourth test case, you can apply superability to the first and the second piles, thus getting a = [1900, 2100, 1600, 3000, 1600].
"""
    Author - Satwik Tiwari .
    19th Jan , 2021  - Tuesday
"""

#===============================================================================================
#importing some useful libraries.

from __future__ import division, print_function
# from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase

# from itertools import *
from heapq import *
from math import gcd, factorial,floor,ceil,sqrt,log2

from copy import deepcopy
from collections import deque


from bisect import bisect_left as bl
from bisect import bisect_right as br
from bisect import bisect

#==============================================================================================
#fast I/O region
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)

# inp = lambda: sys.stdin.readline().rstrip("\r\n")

#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
  def wrapped_func(*args, **kwargs):
    if stack: return f(*args, **kwargs)
    to = f(*args, **kwargs)
    while True:
      if type(to) is GeneratorType:
        stack.append(to)
        to = next(to)
        continue
      stack.pop()
      if not stack: break
      to = stack[-1].send(to)
    return to
  return wrapped_func
#### END ITERATE RECURSION ####

#===============================================================================================
#some shortcuts

def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var))  #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
# def graph(vertex): return [[] for i in range(0,vertex+1)]
def testcase(t):
    for pp in range(t):
        solve(pp)
def google(p):
    print('Case #'+str(p)+': ',end='')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(x, y, p) :
    y%=(p-1)  #not so sure about this. used when y>p-1. if p is prime.
    res = 1     # Initialize result
    x = x % p  # Update x if it is more , than or equal to p
    if (x == 0) :
        return 0
    while (y > 0) :
        if ((y & 1) == 1) : # If y is odd, multiply, x with result
            res = (res * x) % p

        y = y >> 1      # y = y/2
        x = (x * x) % p
    return res
def ncr(n,r): return factorial(n) // (factorial(r) * factorial(max(n - r, 1)))
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
inf = pow(10,20)
mod = 10**9+7
#===============================================================================================
# code here ;))

def do(b,n):
    ans = True
    for i in range(n-1):
        if(i == n-2):
            if(b[i] != b[i+1]):
                ans = False
        else:
            if(b[i] <= b[i+1]):
                b[i+1]-=b[i]
            else:
                ans = False
    return ans

def chck(a):
    ans = False
    b = deepcopy(a)
    n = len(a)
    ans = ans | do(b,n)
    b = deepcopy(a)
    b[0],b[1] = b[1],b[0]
    ans = ans | do(b,n)
    b = deepcopy(a)
    b[n-1],b[n-2] = b[n-2],b[n-1]
    ans = ans | do(b,n)
    return ans

def solve(case):
    n = int(inp())
    b = lis()

    if(n == 2):
        if(b[0] == b[1]):
            print('YES')
        else:
            print('NO')
        return

    if(chck(b)):
        print('YES')
        return

    l = deepcopy(b)
    left = [0]*n
    left[0] = 1
    for i in range(1,n-1):
        if(l[i] >= l[i-1]):
            l[i]-=l[i-1]
            # a[i] = 0
            left[i] = 1 and left[i-1]
        else:
            left[i] = 0
    r = deepcopy(b)
    right = [0]*n
    right[n-1] = 1
    for i in range(n-2,0,-1):
        if(r[i] >= r[i+1]):
            r[i]-=r[i+1]
            right[i] = 1 and right[i+1]
        else:
            right[i] = 0

    # print(l,left)
    # print(r,right)
    ans = False
    for i in range(n-3):
        if(left[i] and right[i+3] and b[i+2] >= l[i] and b[i+1] >= r[i+3] and b[i+2]-l[i] == b[i+1]-r[i+3]):
            ans = True
            # print(i)
    print('YES' if ans else 'NO')



# testcase(1)
testcase(int(inp()))





