human
There are n block towers in a row, where tower i has a height of a_i. You're part of a building crew, and you want to make the buildings look as nice as possible. In a single day, you can perform the following operation:

  * Choose two indices i and j (1 ≤ i, j ≤ n.
i ≠ j), and move a block from tower i to tower j. This essentially decreases a_i by 1 and increases a_j by 1. 


You think the ugliness of the buildings is the height difference between the tallest and shortest buildings. Formally, the ugliness is defined as max(a)-min(a). 

What's the minimum possible ugliness you can achieve, after any number of days?

Input

The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t cases follow.

The first line of each test case contains one integer n (2 ≤ n ≤ 100) — the number of buildings.

The second line of each test case contains n space separated integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^7) — the heights of the buildings.

Output

For each test case, output a single integer — the minimum possible ugliness of the buildings.

Example

Input

3
3
10 10 10
4
3 2 1 2
5
1 2 3 1 5

Output

0
0
1

Note

In the first test case, the ugliness is already 0.

In the second test case, you should do one operation, with i = 1 and j = 3. The new heights will now be [2, 2, 2, 2], with an ugliness of 0.

In the third test case, you may do three operations: 

  1. with i = 3 and j = 1. The new array will now be [2, 2, 2, 1, 5], 
  2. with i = 5 and j = 4. The new array will now be [2, 2, 2, 2, 4], 
  3. with i = 5 and j = 3. The new array will now be [2, 2, 3, 2, 3]. 

The resulting ugliness is 1. It can be proven that this is the minimum possible ugliness for this test.
# imp
Here's the minimum possible ugliness you can achieve, after any number of days?
assistant
#Here's the minimum possible ugliness you can achieve, after any number of days.
ort heapq
# import copy
# from collections import deque
# import sys
# import bisect

# def cost(a, b):
#     return abs(a-b)

# def solve(n, arr):
#     arr.sort()
#     arr = deque(arr)
#     maxh = arr.pop()
#     minh = arr.popleft()
#     prev = abs(maxh-minh)
#     ct = prev
#     ctct = 0
#     if ct < 2:
#         return ct
#     change = True
#     while change:
#         maxh = maxh-ct//2
#         minh = minh+ct//2
#         bisect.insort(arr, maxh)
#         bisect.insort(arr, minh)
#         maxh = arr.pop()
#         minh = arr.popleft()
#         ct = abs(maxh-minh)
#         if ct == prev:
#             if ctct < n:
#                 ctct += 1
#             else:
#                 change = False
#         else:
#             prev = ct
#             ctct = 0

#     return ct

def fsolve(n, arr):
    s = sum(arr)
    if s % n == 0:
        return 0
    else:
        return 1
    # minheap = heapq.heapify(copy(arr))
    # maxheap = [-1*i for i in arr]
    # maxheap = heapq.heapify(copy(maxheap))
    # nochange = False
    # while (minheap[0]-abs(maxheap[0])) != 0 and not nochange:
    #     maxh = abs(heapq.heappop(maxheap))
    #     minh = abs(heapq.heappop(minheap))
    #     cost = maxh-minh
    #     if cost == 0:
    #         return 0
    #     else:
    #         maxh = maxh - cost//2
    #         minh = minh + cost//2
    #         heapq.heappush(maxheap)

if __name__ == "__main__":
    t = int(input())
    for i in range(t):
        n = int(input())
        arr = list(map(int, input().rstrip().split()))
        ans = fsolve(n, arr)
        print(ans)