1552_D. Array Differentiation_66
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.
import sys
import math

if sys.subversion[0] == "PyPy":
    import io, atexit

    sys.stdout = io.BytesIO()
    atexit.register(lambda: sys.__stdout__.write(sys.stdout.getvalue()))

    sys.stdin = io.BytesIO(sys.stdin.read())
    input = lambda: sys.stdin.readline().rstrip()

t = int(raw_input())

# python codesforces210710.py < input.txt
for x in range(t):
    n = int(raw_input())
    str1 = raw_input()
    li = str1.split(' ')
    li = map(int, li)
    tf = 0
    #ssu = sum(li)
    newvec = [-1] * n
    #matri = []
    #newvec = vec[:]
    #matri.append(newvec[:])

    for i in range(1, 3 ** n):
        co = 1
        for j in range(0, n):
            if co == 1 and newvec[j] == 1:
                newvec[j] = -1
                co = 1
            else:
                newvec[j] = newvec[j] + co
                break
        #matri.append(newvec[:])
    # print(matri)


        ssu = [a * b for a, b in zip(newvec, li)]
        if sum(ssu) == 0:
                # print(matri[i])
            tf += 1
            if tf == 2:
                break
    if tf == 2:
        print('YES')
    else:
        print('NO')
