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 <cstdio>
#include <functional>
#include <algorithm>
#include <vector>
#include <limits>
#include <numeric>
#include <type_traits>
#include <iostream>
#include <queue>
#include <bits/stdc++.h>

#define INF (1ll<<60)
#define MOD 1000000007
#define BOUND 27182818284

typedef long long ll;
typedef long long int lli;

#define rep(i,n) for(int i = 0; i < (int)(n); i++)
#define repi(i,a,b) for(int i=int(a);i<int(b);++i)

template<class T>bool umax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool umin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

using namespace std;

// gcd
int gcd(int a, int b){
    if(a==0)
        return b;
    return gcd(b%a,a);
}
// gcd array
int findGCD(int arr[], int n){
    int result = arr[0];
    for (int i=1; i < n; i++)
        result = gcd(arr[i], result);

    return result;
}

// sort big to small
// sort(A,A+N, greater<int>());

// v.front() = -BOUND;
// v.back() = BOUND;

// Iterator
// v.begin
// v.end

int main() {
    long long N,K;;
    long long ans=1;
    cin >> N>>K;

    rep(i,N){
        if(ans+K>2*ans){
            ans*=2;
        }else{
            ans+=K;
        }
    }
    cout << ans << endl;
    return 0;
}