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].
from __future__ import division, print_function
from itertools import permutations 
import threading,bisect,math,heapq,sys
from collections import deque
# threading.stack_size(2**27)
# sys.setrecursionlimit(10**4)
from sys import stdin, stdout
i_m=9223372036854775807    
def cin():
    return map(int,sin().split())
def ain():                           #takes array as input
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input()) 
prime=[]
def dfs(n,d,v):
    v[n]=1
    x=d[n]
    for i in x:
        if i not in v:
            dfs(i,d,v)
    return p 
def block(x): 
      
    v = []  
    while (x > 0): 
        v.append(int(x % 2)) 
        x = int(x / 2) 
    ans=[]
    for i in range(0, len(v)): 
        if (v[i] == 1): 
            ans.append(2**i)  
    return ans 

"""**************************MAIN*****************************"""
def main():
    def ch(a,n):
        c=[-5]*n
        c[0]=a[0]
        for i in range(n-1):
            if a[i+1]>=c[i]:
                c[i+1]=a[i+1]-c[i]
            else:
                break
        return c
    t=inin()
    for _ in range(t):
        n=inin()
        a=[0]
        a.extend(ain())
        a.append(0)
        n+=2
        p=ch(a[:],n)
        s=ch(a[::-1],n)
        s=s[::-1]
        ans="NO"
        if p[-1]==0 or s[0]==0:
            ans="YES"
        # print(p,s,a)
        for i in range(1,n-2):
            if p[i-1]==-5 or s[i+2]==-5:
                continue
            g=[p[i-1],a[i+1],a[i],s[i+2]]
            pp=ch(g[:],4)
            ss=ch(g[::-1],4)
            if pp[-1]==0 or ss[-1]==0:
                ans="YES"
        print(ans)
        
"""***********************************************"""
def intersection(l,r,ll,rr):
    # print(l,r,ll,rr)
    if (ll > r or rr < l): 
            return 0
    else: 
        l = max(l, ll) 
        r = min(r, rr)
    return max(0,r-l+1) 
######## Python 2 and 3 footer by Pajenegod and c1729
fac=[]
def fact(n,mod):
    global fac
    fac.append(1)
    for i in range(1,n+1):
        fac.append((fac[i-1]*i)%mod)
    f=fac[:]
    return f
def nCr(n,r,mod):
    global fac
    x=fac[n]
    y=fac[n-r]
    z=fac[r]
    x=moddiv(x,y,mod)
    return moddiv(x,z,mod)
def moddiv(m,n,p):
    x=pow(n,p-2,p)
    return (m*x)%p
def GCD(x, y): 
    x=abs(x)
    y=abs(y)
    if(min(x,y)==0):
        return max(x,y)
    while(y): 
        x, y = y, x % y 
    return x 
def Divisors(n) : 
    l = []  
    ll=[]
    for i in range(1, int(math.sqrt(n) + 1)) :
        if (n % i == 0) : 
            if (n // i == i) : 
                l.append(i) 
            else : 
                l.append(i)
                ll.append(n//i)
    l.extend(ll[::-1])
    return l
def SieveOfEratosthenes(n): 
    global prime
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    f=[]
    for p in range(2, n): 
        if prime[p]: 
            f.append(p)
    return f
def primeFactors(n): 
    a=[]
    while n % 2 == 0: 
        a.append(2) 
        n = n // 2 
    for i in range(3,int(math.sqrt(n))+1,2):  
        while n % i== 0: 
            a.append(i) 
            n = n // i  
    if n > 2: 
        a.append(n)
    return a
"""*******************************************************"""
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
import os
from io import IOBase, BytesIO
BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
 
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None
 
    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
 
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()
 
    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            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') 
# Cout implemented in Python
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'
 
# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
    conv = ord if py2 else lambda x:x
    A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'R' [0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-' [0]: sign = -1
            elif s[i] != b'\r' [0]:
                A.append(sign*numb)
                numb = zero; sign = 1
            i += 1
    except:pass
    if s and s[-1] >= b'R' [0]:
        A.append(sign*numb)
    return A
 
# threading.Thread(target=main).start()
if __name__== "__main__":
  main()