### Magical game of sum

#### Description:

Yesterday, puppy Tuzik learned a magically efficient method to find the sum of the integers from 1 to N. He denotes it as sum(N). But today, as a true explorer, he defined his own new function: sum(D, N), which means the operation sum applied D times: the first time to N, and each subsequent time to the result of the previous operation. For example, if D = 2 and N = 3, then sum(2, 3) equals to sum(sum(3)) = sum(1 + 2 + 3) = sum(6) = 21. Tuzik wants to calculate some values of the sum(D, N) function. Will you help him with that? Input The first line contains a single integer T, the number of test cases. Each test case is described by a single line containing two integers D and N. Output For each testcase, output one integer on a separate line. Constraints 1 < T< 16 1

#### Program :

#include<stdio.h>

int main()

{

long long int t,d,n;

scanf("%lld",&t);

while(t--)

{

scanf("%lld %lld",&d,&n);

while(d--)

{

n=(n*(n+1))/2;

}

printf("%lld\n",n);

}

return 0;

}