/*
ID: pilania2
TASK: test
LANG: C++
*/
#include "bits/stdc++.h"
using namespace std;
#define int               long long
#define pb                push_back
#define ppb               pop_back
#define pf                push_front
#define ppf               pop_front
#define all(x)            (x).begin(),(x).end()
#define uniq(v)           (v).erase(unique(all(v)),(v).end())
#define sz(x)             (int)((x).size())
#define fr                first
#define sc                second
#define pii               pair<int,int>
#define rep(i,a,b)        for(int i=a;i<b;i++)
#define mem1(a)           memset(a,-1,sizeof(a))
#define mem0(a)           memset(a,0,sizeof(a))
#define ppc               __builtin_popcount
#define ppcll             __builtin_popcountll
 
template<typename T1,typename T2>istream& operator>>(istream& in,pair<T1,T2> &a){in>>a.fr>>a.sc;return in;}
template<typename T1,typename T2>ostream& operator<<(ostream& out,pair<T1,T2> a){out<<a.fr<<" "<<a.sc;return out;}
template<typename T,typename T1>T amax(T &a,T1 b){if(b>a)a=b;return a;}
template<typename T,typename T1>T amin(T &a,T1 b){if(b<a)a=b;return a;}
 
const long long INF=1e18;
const int32_t M=1e9+7;
const int32_t MM=998244353;
const int N=1e9+1;
 
void solve(){
    int a,b,c;
    cin>>a>>b>>c;
    int sum=a+b+c;
    int n=(sum/9);
    if(a<n || b<n || c<n){
        cout<<"NO"<<endl;
        return ;
    }else if(sum%9==0 && a>=n && b>=n && c>=n){
        cout<<"YES"<<endl;
    }else{
        cout<<"NO"<<endl;
    }
    
}
 
signed main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    //ofstream fout (test.out");
    //ifstream fin ("test.in");
    #ifdef SIEVE
        sieve();
    #endif
    #ifdef NCR
        init();
    #endif
    int t=1;
    cin>>t;
    rep(i,0,t) {
        //cout<<"Case #"<<i+1<<": ";
        solve();
    }
    return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

You are playing a new computer game in which you have to fight monsters. In a dungeon you are trying to clear, you met three monsters; the first of them has a health points, the second has b health points, and the third has c.

To kill the monsters, you can use a cannon that, when fired, deals 1 damage to the selected monster. Every 7-th (i. e. shots with numbers 7, 14, 21 etc.) cannon shot is enhanced and deals 1 damage to all monsters, not just one of them. If some monster's current amount of health points is 0, it can't be targeted by a regular shot and does not receive damage from an enhanced shot.

You want to pass the dungeon beautifully, i. e., kill all the monsters with the same enhanced shot (i. e. after some enhanced shot, the health points of each of the monsters should become equal to 0 for the first time). Each shot must hit a monster, i. e. each shot deals damage to at least one monster.

Input

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

Each test case consists of a single line that contains three integers a, b and c (1 ≤ a, b, c ≤ 10^8) — the number of health points each monster has.

Output

For each test case, print YES if you can kill all the monsters with the same enhanced shot. Otherwise, print NO. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).

Example

Input


3
3 2 4
1 1 1
10 1 7


Output


YES
NO
NO

Note

In the first test case, you can do as follows: 1-th shot to the first monster, 2-th shot to the second monster, 3-th shot to the third monster, 4-th shot to the first monster, 5-th shot to the third monster, 6-th shot to the third monster, and 7-th enhanced shot will kill all the monsters.

In the second test case, you can't kill monsters with the same enhanced shot, because the total number of health points of monsters is 3, and you will kill them in the first 3 shots.

### Response

```cpp
/*
ID: pilania2
TASK: test
LANG: C++
*/
#include "bits/stdc++.h"
using namespace std;
#define int               long long
#define pb                push_back
#define ppb               pop_back
#define pf                push_front
#define ppf               pop_front
#define all(x)            (x).begin(),(x).end()
#define uniq(v)           (v).erase(unique(all(v)),(v).end())
#define sz(x)             (int)((x).size())
#define fr                first
#define sc                second
#define pii               pair<int,int>
#define rep(i,a,b)        for(int i=a;i<b;i++)
#define mem1(a)           memset(a,-1,sizeof(a))
#define mem0(a)           memset(a,0,sizeof(a))
#define ppc               __builtin_popcount
#define ppcll             __builtin_popcountll
 
template<typename T1,typename T2>istream& operator>>(istream& in,pair<T1,T2> &a){in>>a.fr>>a.sc;return in;}
template<typename T1,typename T2>ostream& operator<<(ostream& out,pair<T1,T2> a){out<<a.fr<<" "<<a.sc;return out;}
template<typename T,typename T1>T amax(T &a,T1 b){if(b>a)a=b;return a;}
template<typename T,typename T1>T amin(T &a,T1 b){if(b<a)a=b;return a;}
 
const long long INF=1e18;
const int32_t M=1e9+7;
const int32_t MM=998244353;
const int N=1e9+1;
 
void solve(){
    int a,b,c;
    cin>>a>>b>>c;
    int sum=a+b+c;
    int n=(sum/9);
    if(a<n || b<n || c<n){
        cout<<"NO"<<endl;
        return ;
    }else if(sum%9==0 && a>=n && b>=n && c>=n){
        cout<<"YES"<<endl;
    }else{
        cout<<"NO"<<endl;
    }
    
}
 
signed main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    //ofstream fout (test.out");
    //ifstream fin ("test.in");
    #ifdef SIEVE
        sieve();
    #endif
    #ifdef NCR
        init();
    #endif
    int t=1;
    cin>>t;
    rep(i,0,t) {
        //cout<<"Case #"<<i+1<<": ";
        solve();
    }
    return 0;
}

```