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.
Solve the task in PYTHON3.
assistant
#Here's the minimum possible ugliness you can achieve, after any number of days.

#!/usr/bin/env python

# जय श्री गणेश
#--MODULES BEGINS--
from __future__ import division, print_function
import sys
'''--python 2 and pypy begins--  - comment this line to run the below code
if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip
# --python 2 and pypy ends--'''
import io, os, re, time, random
from io import BytesIO, IOBase
import math, functools, collections
# import bisect, heapq, numpy
# from contextlib import contextmanager
# from jinja2 import Template
#--MODULES ENDS--

#--SHORTHANDS BEGINS--
# input = raw_input
# input = sys.stdin.readline
# sys.stdout.flush()
# alphabets = "abcdefghijklmnopqrstuvwxyz"
# from statistics import mode
# from functools import reduce
# from bisect import bisect
# from collections import defaultdict
# from collections import Counter
# hp = defaultdict(lambda: 0)
# hp = collections.defaultdict(int, collections.Counter(lis))
# lis = sorted(hp.items(), key = lambda kv:(kv[1], kv[0]))
# lis = sorted(lis, key = lambda ele: (ele[1], ele[0]))
# lis = list(filter(lambda a: a[1] != min(hp.values()), lis))
# input = lambda: sys.stdin.readline().strip()
# imap = lambda: map(int,input().split())
# ilist = lambda: list(map(int, input().split()))
# sys.stdout.write(str(ans) + "\n")
# for line in sys.stdin
# except: print(f'Unknown Error: {sys.exc_info()[1]}')
# w = int(input()); print("YNEOS"[(w > 2 and not w % 2)::2]);

# return self.inorderTraversal(root.left) + [root.val] + self.inorderTraversal(root.right) if root else list()
#--SHORTHANDS ENDS--

#----    CODE BEGINS    ----->

#--GLOBAL VALUES BEGINS--
class ConstantMod:
    ''' Constant Values for Competitive Programming '''
    # __slots__ = ['_modd', '_maxx', ]  # for fixed number of attributes in __dict__ of the class's instance object
    def __init__(self, n = None, *args, **kwargs):
        self._modd = n
        # self.minn = -(n)
        self._maxx = kwargs['maxvalue'] if 'maxvalue' in kwargs else int(1e9 + 7)
       ...