Problem:
Score : 200 points

Problem Statement

Square1001 has seen an electric bulletin board displaying the integer 1.
He can perform the following operations A and B to change this value:

Operation A: The displayed value is doubled.

Operation B: The displayed value increases by K.

Square1001 needs to perform these operations N times in total.
Find the minimum possible value displayed in the board after N operations.

Constraints

1 \leq N, K \leq 10

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
K

Output

Print the minimum possible value displayed in the board after N operations.

Sample Input 1

4
3

Sample Output 1

10

The value will be minimized when the operations are performed in the following order: A, A, B, B.

In this case, the value will change as follows: 1 → 2 → 4 → 7 → 10.

Sample Input 2

10
10

Sample Output 2

76

The value will be minimized when the operations are performed in the following order: A, A, A, A, B, B, B, B, B, B.

In this case, the value will change as follows: 1 → 2 → 4 → 8 → 16 → 26 → 36 → 46 → 56 → 66 → 76.

By the way, this contest is AtCoder Beginner Contest 076.

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>                                                                                                                                         
                                                                                                                                                           
int main(void)                                                                                                                                             
{                                                                                                                                                          
        int n, k, i, now = 1;                                                                                                                              
                                                                                                                                                           
        scanf("%d", &n);                                                                                                                                   
        scanf("%d", &k);                                                                                                                                   
                                                                                                                                                           
        for (i = 0; i < n; i++) {                                                                                                                          
                if (now * 2 < now + k) {                                                                                                                   
                        now = now * 2;                                                                                                                     
                } else {                                                                                                                                   
                        now = now + k;                                                                                                                     
                }                                                                                                                                          
        }                                                                                                                                                  
        printf("%d\n", now);                                                                                                                               
        return 0;                                                                                                                                          
}