human
Give me a the Python program you requested given the following:
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.
from
assistant
#Here is the Python program you requested.
math import *
from decimal import *
from decimal import Decimal as D
from string import ascii_lowercase as lc
from string import ascii_uppercase as uc
from itertools import *
from itertools import permutations as perm
from itertools import product as prod
from sys import *
ip = input
ra = lambda: list(map(int, input().split()))
ri = lambda: int(input())
rd = lambda: map(int, input().split())
MOD = 10 ** 9 + 7
MOD2 = 998244353
def debug(*a):
    stderr.write(' '.join(map(str, a)) + '\n')
def fp(x, p, m):
    r = 1
    while p:
        if p % 2:
            r = r * x % m
            p -= 1
        else:
            x = x * x % m
            p //= 2
    return r
def fact(x, m):
    r = 1
    for i in range(2, x + 1):
        r = r * i % m
    return r
def C(n, k, m):
    return fact(n, m) * fp(fact(n - k, m), m - 2, m) % m * fp(fact(k, m), m - 2, m) % m
class SegmentTree:
    def f(self, a, b):
        return a + b
    def __init__(self, n, a):
        self.tree = [0] * (2 * n)
        self.n = n
        for i in range(n):
            self.tree[n + i] = a[i]
        for i in range(n - 1, 0, -1):
            self.tree[i] = self.f(self.tree[2 * i], self.tree[2 * i + 1])
    def q(self, l, r):
        l += self.n
        r += self.n
        res = 0
        while r > 1:
            if l % 2:
                res = self.f(res, self.tree[l]) 
                l += 1
            if r % 2:
                r -= 1
                res = self.f(res, self.tree[r])
            r //= 2
            l //= 2
        return res
    def u(self, index, value):
        index += self.n
        self.tree[index] = value
        index //= 2
        while index > 0:
            self.tree[index] = self.f(self.tree[2 * index], self.tree[2 * index + 1])
            index //= 2
def z_function(s):
    l, r, n = 0, 0, len(s)
    z = [0] * n
    z[0] = 0
    for i in range(1, n):
        if r >= i: z[i] = min(z[i - l], r - i + 1)
        while z[i] + i < n and s[z[i]] == s[z[i] + i]: z[i] += 1
        if i + z[i] - 1 > r:
            l = i
       ...