human
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.":


 ... What is a the Python code you requested?
assistant
#Here is the Python code you requested.
 def find_common(a1, a2):
    list_len = len(a1)
    list_len2 = len(a2)
    a3 = []
    iz = jz = 0
    while iz < list_len and jz < list_len2:
        if a1[iz] < a2[jz]:
            iz += 1
        elif a2[jz] < a1[iz]:
            jz += 1
        else:
            a3.append(a1[iz])    
            iz +=1
            jz +=1
    return a3

t=int(input())
for i in range(t):
    y=0
    n=int(input())
    a=list(map(int,input().split()))
    if 0 in a:
        print("YES")
    else:
        if n<=5:
            for j in range(1,3**n):
                tz=0
                for k in range(n):
                    if (j//3**k)%3==1:
                        tz+=a[k]
                    elif (j//3**k)%3==2:
                        tz-=a[k]
                if tz==0:
                    y=1
                    break
            if y==0:
                print("NO")
            else:
                print("YES")
        else:
            l=[]
            for j in range(1,3**5):
                tz=0
                for k in range(5):
                    if (j//3**k)%3==1:
                        tz+=a[k]
                    elif (j//3**k)%3==2:
                        tz-=a[k]
                l.append(abs(tz))
            l=list(set(l))
            l.sort()
            a=a[5:]
            ln=[]
            for j in range(1,3**len(a)):
                tz=0
                for k in range(len(a)):
                    if (j//3**k)%3==1:
                        tz+=a[k]
                    elif (j//3**k)%3==2:
                        tz-=a[k]
                ln.append(tz)
            ln=list(set(ln))
            ln.sort()
            if find_common(l,ln):
                print("YES")
            elif 0 in ln:
                print("YES")
            elif 0 in l:
                print("YES")
            else:
                print("NO")