human
Can you solve the below in PYTHON3?
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.
assistant
#Here is the Python code you requested.

def naiveSolve():
    
    
    
    return


def main():
    
    t=int(input())
    allans=[]
    for _ in range(t):
        n=int(input())
        a=readIntArr()
        
        
        ans='NO'
        # as long as sum of some of a is sum of some others in a, it is possible.
        for mask1 in range(1,1<<n): # non-empty
            summ1=0
            for i in range(n):
                if (mask1&(1<<i))>0:
                    summ1+=a[i]
            if summ1==0: # say we have a=1,2,-3. we need 0,1,3,0. But we already had 0 so we don't need an extra 0.
                ans='YES'
            for mask2 in range(1,1<<n): # non-empty
                if mask1&mask2>0:
                    continue
                summ2=0
                for i in range(n):
                    if (mask2&(1<<i))>0:
                        summ2+=a[i]
                if summ1==summ2: # mask1's last b will intersect with mask2's last b
                    ans='YES'
        allans.append(ans)
    multiLineArrayPrint(allans)
    
    return


import sys
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
# input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.

def oneLineArrayPrint(arr):
    print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
    print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
    print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
 
def readIntArr():
    return [int(x) for x in input().split()]
# def readFloatArr():
#     return [float(x) for x in input().split()]
 
def makeArr(defaultValFactory,dimensionArr): # eg. makeArr(lambda:0,[n,m])
    dv=defaultValFactory;da=dimensionArr
    if len(da)==1:return [dv() for _ in range(da[0])]
    else:return [makeArr(dv,da[1:]) for _ in range(da[0])]
 
def queryInteractive(l,r):
    print('? {} {}'.format(l,r))
    sys.stdout.flush()
    return int(input())
 
def answerInteractive(x):
    print('! {}'.format(x))
    sys.stdout.flush()
...