You are given a sequence of n integers a_1,   a_2,   ...,   a_n.

Does there exist a sequence of n integers b_1,   b_2,   ...,   b_n such that the following property holds?

  * For each 1 ≤ i ≤ n, there exist two (not necessarily distinct) indices j and k (1 ≤ j,   k ≤ n) such that a_i = b_j - b_k. 

Input

The first line contains a single integer t (1 ≤ t ≤ 20) — the number of test cases. Then t test cases follow.

The first line of each test case contains one integer n (1 ≤ n ≤ 10).

The second line of each test case contains the n integers a_1,   ...,   a_n (-10^5 ≤ a_i ≤ 10^5).

Output

For each test case, output a line containing YES if a sequence b_1,   ...,   b_n satisfying the required property exists, and NO otherwise.

Example

Input


5
5
4 -7 -1 5 10
1
0
3
1 10 100
4
-3 2 10 2
9
25 -171 250 174 152 242 100 -205 -258


Output


YES
YES
NO
YES
YES

Note

In the first test case, the sequence b = [-9,   2,   1,   3,   -2] satisfies the property. Indeed, the following holds: 

  * a_1 = 4 = 2 - (-2) = b_2 - b_5; 
  * a_2 = -7 = -9 - (-2) = b_1 - b_5; 
  * a_3 = -1 = 1 - 2 = b_3 - b_2; 
  * a_4 = 5 = 3 - (-2) = b_4 - b_5; 
  * a_5 = 10 = 1 - (-9) = b_3 - b_1. 



In the second test case, it is sufficient to choose b = [0], since a_1 = 0 = 0 - 0 = b_1 - b_1.

In the third test case, it is possible to show that no sequence b of length 3 satisfies the property.
/** * * * * * * * * * * * * **\
 *                           *
 *  Author: Haidara Nassour  *
 *        LANG: C++17        *
 *                           *
\** * * * * * * * * * * * * **/
#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define int long long
#define rep(i,x,n) for(int i=x;i<(n);i++)
#define FOR(i,n) rep(i,0,n)
#define per(i,x,n) for(int i=x;i>(n);i--)
#define ROF(i,x) for(int i=x;i>=0;i--)
#define v(i) vector< i >
#define p(i,j) pair< i , j >
#define pii pair<int,int>
#define m(i,j) map< i , j >
#define um(i,j) unordered_map< i , j >
#define max_heap(i) priority_queue< i >
#define min_heap(i) priority_queue< i , vector< i > ,greater< i > >
#define ff first
#define all(x) x.begin(),x.end()
#define ss second
#define pp push_back
#define minimize(x,y) x=min(x,y);
#define maximize(x,y) x=max(x,y);
#define debug(x) cout<<#x<<" "<<x;
using namespace std;
using namespace __gnu_pbds;
void SIO(string name="")
{
    if(name!="")
    {
        freopen((name+".in").c_str(),"r",stdin);
        freopen((name+".out").c_str(),"w",stdout);
    }
}
template <class T> using o_set = tree<T, null_type, less<T>,rb_tree_tag, tree_order_statistics_node_update>;
///order_of_key = find index of element x
///find_by_order = find value at index x
const int inf=1LL<<62LL;
const int mod=1e9+7;
const int maxn=11;
int n;
v(int)a(maxn);
string ans;
v(int)b;
void solve(int inx=1)
{
    if(inx==n)
    {
        FOR(i,n)
        rep(j,i,n)
        if(abs(b[i]-b[j])==abs(a[0]))
            ans="YES";
        return ;
    }
    FOR(i,b.size())
    {
        b.pp(b[i]-a[inx]);
        solve(inx+1);
        b.pop_back();
    }
    return ;
}
signed main()
{
    SIO("");
    int t=1;
    cin>>t;
    while(t--)
    {
        ans="NO";
        cin>>n;
        FOR(i,n)
        cin>>a[i];
        int mx=-inf;
        int inx=0;
        FOR(i,n)
        {
            b.clear();
            b.pp(a[i]);
            solve();
        }
        cout<<ans<<endl;
    }
}
