Margarite and the best present

Little girl Margarita is a big fan of competitive programming. She especially loves problems about arrays and queries on them.

Recently,she was presented with an array?aa?of the size of?109109?elements that is filled as follows:

a1=?1
a2=2
a3=?3
a4=4
a5=?5
And so on ...
That is,the value of the?ii-th element of the array?aa?is calculated using the formula?ai=i?(?1)iai=i?(?1)i.

She immediately came up with?qq?queries on this array. Each query is described with two numbers:?ll?and?rr. The answer to a query is the sum of all the elements of the array at positions from?ll?to?rr?inclusive.

Margarita really wants to know the answer to each of the requests. She doesn‘t want to count all this manually,but unfortunately,she couldn‘t write the program that solves the problem either. She has turned to you?— the best programmer.

Help her find the answers!

Input

The first line contains a single integer?qq?(1≤q≤1031≤q≤103)?— the number of the queries.

Each of the next?qq?lines contains two integers?ll?and?rr?(1≤l≤r≤1091≤l≤r≤109)?— the descriptions of the queries.

Output

Print?qq?lines,each containing one number?— the answer to the query.

Example

input

Copy

5
1 3
2 5
5 5
4 4
2 3
output

Copy

-2
-2
-5
4
-1
Note

In the first query,you need to find the sum of the elements of the array from position?11?to position?33. The sum is equal to?a1+a2+a3=?1+2?3=?2a1+a2+a3=?1+2?3=?2.

In the second query,you need to find the sum of the elements of the array from position?22?to position?55. The sum is equal to?a2+a3+a4+a5=2?3+4?5=?2a2+a3+a4+a5=2?3+4?5=?2.

In the third query,you need to find the sum of the elements of the array from position?55?to position?55. The sum is equal to?a5=?5a5=?5.

In the fourth query,you need to find the sum of the elements of the array from position?44?to position?44. The sum is equal to?a4=4a4=4.

In the fifth query,you need to find the sum of the elements of the array from position?22?to position?33. The sum is equal to?a2+a3=2?3=?1a2+a3=2?3=?1.

题解：对于多种情况考虑分析即可

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作者：black_hole6
来源：CSDN
原文：https://blog.csdn.net/lbperfect123/article/details/84449954
版权声明：本文为博主原创文章，转载请附上博文链接！

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
using namespace std;
 
int main()
{
    int n;
    cin>>n;
 
    for(int t=0;t<n;t++)
    {
        int a,b;
        long long int sum;
        scanf("%d%d",&a,&b);
        if(a%2==1&&b%2==1)
        {
            sum=-1*((a-b)/2+b);
        }
        else if(a%2==1&&b%2==0)
        {
            sum=(b-a+1)/2;
        }
        else if(a%2==0&&b%2==1)
        {
            sum=-1*(b-a+1)/2;
        }
        else
        {
            sum=(a-b)/2+b;
        }
        printf("%lldn",sum);
        
    }
    return 0;
 }