A. Collecting Coins

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Polycarp has three sisters: Alice, Barbara, and Cerene. They're collecting coins. Currently, Alice has aa coins, Barbara has bb coins and Cerene has cc coins. Recently Polycarp has returned from the trip around the world and brought nn coins.

He wants to distribute all these nn 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 AA coins to Alice, BB coins to Barbara and CC coins to Cerene (A+B+C=nA+B+C=n), then a+A=b+B=c+Ca+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 nn coins between sisters in a way described above.

You have to answer tt independent test cases.

Input

The first line of the input contains one integer tt (1≤t≤1041≤t≤104) — the number of test cases.

The next tt lines describe test cases. Each test case is given on a new line and consists of four space-separated integers a,b,ca,b,c and nn (1≤a,b,c,n≤1081≤a,b,c,n≤108) — 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 nn coins between his sisters and "NO" otherwise.

Example

input

Copy

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

output

Copy

YES
YES
NO
NO
YES

解题说明：此题是一道模拟题，题意是你有三个姐妹她们分别有 a , b , c枚硬币，你有n枚，你可以把硬币随意分给她们（必须分完），使她们的硬币数A = B = C 可以求出x=(a+b+c+n-3*a)/3,y=x=(a+b+c+n-3*b)/3,z=(a+b+c+n-3*c)/3,如果三个式子的分母只要有一个小于零或者不能被3整除就不能分完，否则可以。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>

using namespace std;

int main()
{
	int t, a, b, c, n, k, i;
	scanf("%d", &t);
	for (i = 0; i < t; i++)
	{
		scanf("%d %d %d %d", &a, &b, &c, &n);
		k = n + a + c - 2 * b;
		if (k % 3 == 0 && k / 3 >= 0 && b - a + k / 3 >= 0 && b - c + k / 3 >= 0)
		{
			printf("YES\n");
		}
		else
		{
			printf("NO\n");
		}
	}
	return 0;
}