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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef unsigned long long ull; 
typedef vector<int> vi;
typedef pair<int,int> pi;

#define F first
#define S second
#define PB push_back
#define MP make_pair
#define FOR(i,a,b) for (int i = a; i < b; i++)
#define SQ(a) (a)*(a)

template <class T, class U> ostream& operator<< (ostream &out, pair<T,U> c) { out << "(" << c.first << " "; out << c.second << ")"; return out; }
template <class T> ostream& operator<< (ostream &out, vector<T> c) { for (auto it = c.begin(); it != c.end(); ++it) { if (it != c.begin()) out << " "; out << *it; } return out; }
template <class T> ostream& operator<< (ostream &out, set<T> c) { for (auto it = c.begin(); it != c.end(); ++it) { if (it != c.begin()) out << " "; out << *it; } return out; }
const double PI  = 3.141592653589793238463;


int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n, k, m = 1;
    cin >> n >> k;
    while(n > 0 && m <= k){
        m *= 2;
        n--;
    }
    FOR(i,0,n){
        m += k;
    }
    cout << m << endl;
}