**Factor of 3 Codevita 9 Solutions**

Given an array arr, of size N, find whether it is possible to rearrange the elements of array such that sum of no two adjacent elements is divisible by 3.

**Constraints**

1 <= T <= 10

2 <= N <= 10^5

1 <= arr[i] <= 10^5

**Input**

First line contains integer T denoting the number of testcases. Each test cases consists of 2 lines as follows- First line contains integer N denoting the size of the array. Second line contains N space separated integers.

**Output**

For each test case print either “Yes” or “No” (without quotes) on new line.

**Time Limit**

1

**Examples**

**Example 1****Input**

1

4

1 2 3 3**Output**

Yes

**Explanation**

Some of the rearrangements can be {2,1,3,3}, {3,3,1,2}, {2,3,3,1}, {1,3,3,2},… We can see that there exist at least 1 combination {3,2,3,1} where

sum of 2 adjacent number is not divisible by 3. Other combinations

can be {1,3,2,3}, {2,3,1,3}. Hence the output is Yes.

**Example 2****Input**

1

4

3 6 1 9**Output**

No

**Explanation**

All possible combination of {3,6,1,9} are

{1,3,6,9}, {1,3,9,6}, {1,6,9,3}, {1,6,3,9}, {1,9,3,6}, {1,9,6,3}, {6,1,3,9}, {6,1, 9,3}, {6,3,1,9}, {6,3,9,1}, {6,9,1,3}, {6,9,3,1}, {3,1,6,9}, {3,1,9,6}, {3,9,1,6}, {3,9,6,1}, {3,6,1,9}, {3,6,9,1}, {9,1,3,6}, {9,1,6,3}, {9,3,1,6}, {9,3,6,1}, {9,6,1,3}, {9,6,3,1}.

Since none of these combinations satisfy the condition, the output is No

**SOLUTION**

** Program**: Factor of 3 Codevita 9 Solutions in Python

```
for i in range (int(input("Enter Test Case"))):
N = int(input())
list1 = list(map(int,input().split()))
array = []
for i in range(N):
array.append(list1[i]%3)
x = array.count(0)
y = array.count(1)
z = array.count(2)
if x == 0 and y != 0 and z !=0:
print("NO")
elif x == 0 and y == 0 and z !=0:
print("YES")
elif x == 0 and y != 0 and z ==0:
print("YES")
elif x<=(z+y):
print("YES")
else:
print("No")
```

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