codeforces451C

Predict Outcome of the Game

 CodeForces - 451C 

There are n games in a football tournament. Three teams are participating in it. Currently k games had already been played.

You are an avid football fan, but recently you missed the whole k games. Fortunately, you remember a guess of your friend for these k games. Your friend did not tell exact number of wins of each team, instead he thought that absolute difference between number of wins of first and second team will be d₁ and that of between second and third team will be d₂.

You don't want any of team win the tournament, that is each team should have the same number of wins after n games. That's why you want to know: does there exist a valid tournament satisfying the friend's guess such that no team will win this tournament?

Note that outcome of a match can not be a draw, it has to be either win or loss.

Input

The first line of the input contains a single integer corresponding to number of test cases t (1 ≤ t ≤ 10⁵).

Each of the next t lines will contain four space-separated integers n, k, d₁, d₂ (1 ≤ n ≤ 10¹²; 0 ≤ k ≤ n; 0 ≤ d₁, d₂ ≤ k) — data for the current test case.

Output

For each test case, output a single line containing either "yes" if it is possible to have no winner of tournament, or "no" otherwise (without quotes).

Examples

Input

5
3 0 0 0
3 3 0 0
6 4 1 0
6 3 3 0
3 3 3 2

Output

yes
yes
yes
no
no

Note

Sample 1. There has not been any match up to now (k = 0, d₁ = 0, d₂ = 0). If there will be three matches (1-2, 2-3, 3-1) and each team wins once, then at the end each team will have 1 win.

Sample 2. You missed all the games (k = 3). As d₁ = 0 and d₂ = 0, and there is a way to play three games with no winner of tournament (described in the previous sample), the answer is "yes".

Sample 3. You missed 4 matches, and d₁ = 1, d₂ = 0. These four matches can be: 1-2 (win 2), 1-3 (win 3), 1-2 (win 1), 1-3 (win 1). Currently the first team has 2 wins, the second team has 1 win, the third team has 1 win. Two remaining matches can be: 1-2 (win 2), 1-3 (win 3). In the end all the teams have equal number of wins (2 wins).

sol：大力分类讨论四种情况，要及其仔细，否则会挂的很惨。。。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
inline ll read()
{
    ll s=0;
    bool f=0;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<3)+(s<<1)+(ch^48); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0)
    {
        putchar('-'); x=-x;
    }
    if(x<10)
    {
        putchar(x+'0');    return;
    }
    write(x/10);
    putchar((x%10)+'0');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('
')
ll T,n,m,d1,d2;
#define PYES {puts("yes"); continue;}
#define PNO {puts("no"); continue;}
int main()
{
    R(T);
    while(T--)
    {
        R(n); R(m); R(d1); R(d2);
        if(n%3) PNO;
        ll tmp=n/3,oo;
        if(max(d1,d2)>tmp) PNO
        if(d1>=d2) //0,d1,d1-d2
        {
            oo=m-d1-d1+d2;
            if((d1+oo/3)<=tmp&&oo>=0&&oo%3==0) PYES
        }
        else //d2-d1,d2,0
        {
            oo=m-d2-d2+d1;
            if((d2+oo/3)<=tmp&&oo>=0&&oo%3==0) PYES
        }
        oo=m-d1-d2; //d1,0,d2
        if((max(d1,d2)+oo/3)<=tmp&&oo>=0&&oo%3==0) PYES
        oo=m-d1-d2-d2; //d1+d2,d2,0
        if((d1+d2+oo/3)<=tmp&&oo>=0&&oo%3==0) PYES
        oo=m-d1-d1-d2; //0,d1,d1+d2
        if((d1+d2+oo/3)<=tmp&&oo>=0&&oo%3==0) PYES
        PNO
    }
    return 0;
}
/*
Input
8
3 0 0 0
3 3 0 0
6 4 1 0
6 3 3 0
3 3 3 2
3 1 1 1
9 4 1 2
3 3 1 1
Output
yes
yes
yes
no
no
yes
yes
no
*/