You are given a number n and 2*n numbers. Write a program to check whether the sum of the odd numbers is equal to the sum of the even n numbers. The first number is considered odd, the next even, the next odd again, etc. Print as result “Yes” or “No”. In case of yes, print also the sum. In case of no, print also the difference between the odd and the even sums.

### Input

The input data should be read from the console.

• The first line holds an integer n – the count of numbers.
• Each of the next 2*n lines holds exactly one number.

The input data will always be valid and in the format described. There is no need to check it explicitly.

### Output

• The output must be printed on the console.
• Print “Yes, sum=S” where S is the sum of the odd n numbers in case of the sum of the odd n numbers is equal to the sum of the even n numbers.
• Otherwise print “No, diff=D” where D is the difference between the sum of the odd n numbers and the sum of the even n numbers. D should always be a positive number.

### Constraints

• The number n is integer in range [0...500].
• All other numbers are integers in range [-500 000 ... 500 000].
• Allowed working time for your program: 0.25 seconds.
• Allowed memory: 16 MB.

### Examples   edited

Here's my solution, mate:

```using System;

class OddEvenSum
{
static void Main()
{

int[] array = new int[n * 2];

for (int i = 0; i < array.Length; i++)
{
}

int sumEven = 0;
int sumOdd = 0;
for (int i = 0; i < array.Length; i++)
{
if (i % 2 != 0)
{
sumOdd += array[i];
}
else if (i % 2 == 0)
{
sumEven += array[i];
}
}

int dif = Math.Abs(sumEven - sumOdd);
if (sumEven == sumOdd)
{
Console.WriteLine("Yes, sum={0}", sumOdd);
}

else
{
Console.WriteLine("No, diff={0}", dif);
}
}
}
```