human
Give me a the Python program you requested given the following:
PYTHON3 solution for "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 program you requested.
 def r(a, suma, cerca):
    if suma == cerca: return True
    if not a: return False
    last = a.pop()
    possible = (r(a, suma+last, cerca) or r(a, suma-last, cerca)
                or r(a, suma, cerca))
    a.append(last)
    return possible

for _ in range(int(input())):
    n = int(input())
    a = list(map(int, input().split()))
    print("YES" if any(r(a[:i]+a[i+1:], 0, a[i]) for i in range(n)) else "NO")