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.
///..............*****.................///
///   Saiful Alam                      ///
///   Department of CSE                ///
///   Comilla University , Bangladesh. ///
///..............****..................///
#include<bits/stdc++.h>
#define  p5         100005
#define  p6         1000007
#define  PI         acos(-1)
#define  M          1000000007
#define  sc         scanf
#define  ss         second
#define  ff         first
#define  pf         printf
#define  scin(x)    sc("%d",&x)
#define  scin2(x,y) sc("%d%d",&x,&y)
#define  READ(f)    freopen(f, "r", stdin)
#define  WRITE(f)   freopen(f, "w", stdout)
#define  ll         long long
#define  pp         pair <ll,ll>
#define  pq         priority_queue
#define  pb         push_back
#define  mpa         make_pair
#define  yes        pf("YES\n")
#define  no         pf("NO\n")
#define all(v)      v.begin(), v.end()
#define unq(v)      v.erase(unique(all(v)),v.end());
#define valid(x,y) x>=0 && y>=0 && x<n && y<m
//int month[]={31,28,31,30,31,30,31,31,30,31,30,31};
//int dx[5] = {1, -1, 0, 0 };
//int dy[5] = {0, 0, 1, -1};
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define ordered_set tree<ll, null_type,less_equal<ll>, rb_tree_tag,tree_order_statistics_node_update>
using namespace std;

int main()
{
    ios_base::sync_with_stdio(false);
    ll ca;
    cin>>ca;
    while(ca--)
    {
        ll n,i,j,k,cnt=0;
        cin>>n;
        ll a[n+4];
        for(i=0; i<n; i++)
        {
            cin>>a[i];
            if(a[i]<=0) cnt++;
        }
        if(cnt) yes;
        else no;
    }

    return 0;
}


