Problem: Polycarp has three sisters: Alice, Barbara, and Cerene. They're collecting coins. Currently, Alice has a coins, Barbara has b coins and Cerene has c coins. Recently Polycarp has returned from the trip around the world and brought n coins.

He wants to distribute all these n coins between his sisters in such a way that the number of coins Alice has is equal to the number of coins Barbara has and is equal to the number of coins Cerene has. In other words, if Polycarp gives A coins to Alice, B coins to Barbara and C coins to Cerene (A+B+C=n), then a + A = b + B = c + C.

Note that A, B or C (the number of coins Polycarp gives to Alice, Barbara and Cerene correspondingly) can be 0.

Your task is to find out if it is possible to distribute all n coins between sisters in a way described above.

You have to answer t independent test cases.

Input

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

The next t lines describe test cases. Each test case is given on a new line and consists of four space-separated integers a, b, c and n (1 ≤ a, b, c, n ≤ 10^8) — the number of coins Alice has, the number of coins Barbara has, the number of coins Cerene has and the number of coins Polycarp has.

Output

For each test case, print "YES" if Polycarp can distribute all n coins between his sisters and "NO" otherwise.

Example

Input


5
5 3 2 8
100 101 102 105
3 2 1 100000000
10 20 15 14
101 101 101 3


Output


YES
YES
NO
NO
YES

Solution (in PYTHON3):

Solution:
"""
Author    : co_devil Chirag Garg
Institute : JIIT
"""
from math import *
from sys import stdin, stdout
import itertools
import os
import sys
import threading
from collections import deque, Counter, OrderedDict, defaultdict
from heapq import *
# from math import ceil, floor, log, sqrt, factorial, pow, pi, gcd
# from bisect import bisect_left,bisect_right
# from decimal import *,threading
from fractions import Fraction
mod = int(pow(10, 9)+7)


def ii(): return int(input())


def si(): return str(input())


def mi(): return map(int, input().split())


def li(): return list(mi())


def fii(): return int(stdin.readline())


def fsi(): return str(stdin.readline())


def fmi(): return map(int, stdin.readline().split())


def fli(): return list(fmi())


abd = {'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12,
       'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24,
       'z': 25}


def getKey(item): return item[0]


def sort2(l): return sorted(l, key=getKey)


def d2(n, m, num): return [[num for x in range(m)] for y in range(n)]


def isPowerOfTwo(x): return (x and (not (x & (x - 1))))


def decimalToBinary(n): return bin(n).replace("0b", "")


def ntl(n): return [int(i) for i in str(n)]


def powerMod(x, y, p):
    res = 1
    x %= p
    while y > 0:
        if y & 1:
            res = (res * x) % p
        y = y >> 1
        x = (x * x) % p
    return res


graph = defaultdict(list)
visited = [0] * 1000000
col = [-1] * 1000000


def bfs(d, v):
    q = []
    q.append(v)
    visited[v] = 1
    while len(q) != 0:
        x = q[0]
        q.pop(0)
        for i in d[x]:
            if visited[i] != 1:
                visited[i] = 1
                q.append(i)
        print(x)


def make_graph(e):
    d = {}
    for i in range(e):
        x, y = mi()
        if x not in d:
            d[x] = [y]
        else:
            d[x].append(y)
        if y not in d:
            d[y] = [x]
        else:
            d[y].append(x)
    return d


def gr2(n):
    d = defaultdict(list)
    for i in range(n):
        x, y = mi()
        d[x].append(y)
    return d


def connected_components(graph):
    seen = set()

    def dfs(v):
        vs = set([v])
        component = []
        while vs:
            v = vs.pop()
            seen.add(v)
            vs |= set(graph[v]) - seen
            component.append(v)
        return component

    ans = []
    for v in graph:
        if v not in seen:
            d = dfs(v)
            ans.append(d)
    return ans


def primeFactors(n):
    s = set()
    while n % 2 == 0:
        s.add(2)
        n = n // 2
    for i in range(3, int(sqrt(n)) + 1, 2):
        while n % i == 0:
            s.add(i)
            n = n // i
    if n > 2:
        s.add(n)
    return s


def find_all(a_str, sub):
    start = 0
    while True:
        start = a_str.find(sub, start)
        if start == -1:
            return
        yield start
        start += len(sub)


def SieveOfEratosthenes(n, isPrime):
    isPrime[0] = isPrime[1] = False
    for i in range(2, n):
        isPrime[i] = True
    p = 2
    while (p * p <= n):
        if (isPrime[p] == True):
            i = p * p
            while (i <= n):
                isPrime[i] = False
                i += p
        p += 1
    return isPrime


def dijkstra(edges, f, t):
    g = defaultdict(list)
    for l, r, c in edges:
        g[l].append((c, r))

    q, seen, mins = [(0, f, ())], set(), {f: 0}
    while q:
        (cost, v1, path) = heappop(q)
        if v1 not in seen:
            seen.add(v1)
            path = (v1, path)
            if v1 == t:
                return (cost, path)

            for c, v2 in g.get(v1, ()):
                if v2 in seen:
                    continue
                prev = mins.get(v2, None)
                next = cost + c
                if prev is None or next < prev:
                    mins[v2] = next
                    heappush(q, (next, v2, path))
    return float("inf")


def binsearch(a, l, r, x):
    while l <= r:
        mid = l + (r-1)//2
        print(mid)
        if a[mid]:
            return mid
        elif a[mid] > x:
            l = mid-1
        else:
            r = mid+1
    return -1


t = ii()
for _ in range(t):
    a, b, c, n = mi()
    if(a+b+c+n) % 3 == 0:
        k = (a+b+c+n)//3
        if k-a < 0 or k-b < 0 or k-c < 0:
            print("NO")
        else:
            print("YES")
    else:
        print("NO")