Hello 2018

A. Modular Exponentiation

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The following problem is well-known: given integers n and m, calculate

,

where 2ⁿ = 2·2·...·2 (n factors), and denotes the remainder of division of x by y.

You are asked to solve the "reverse" problem. Given integers n and m, calculate

.

Input

The first line contains a single integer n (1 ≤ n ≤ 10⁸).

The second line contains a single integer m (1 ≤ m ≤ 10⁸).

Output

Output a single integer — the value of .

Examples

Input

4
42

Output

10

Input

1
58

Output

0

Input

98765432
23456789

Output

23456789

Note

In the first example, the remainder of division of 42 by 2⁴ = 16 is equal to 10.

In the second example, 58 is divisible by 2¹ = 2 without remainder, and the answer is 0.

特判一下，

然后就直接模了；

#include <iostream>
#define ll long long int
using namespace std;
ll n,m;

int main(){
    cin>>n>>m;
    ll sum = 1;
    if(n>=27){
        cout<<m<<endl;
    }else{
        while(n--){
            sum*=2;
        }
        cout<<m%sum<<endl;
    }
    return 0;
}

B. Christmas Spruce

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Consider a rooted tree. A rooted tree has one special vertex called the root. All edges are directed from the root. Vertex u is called a child of vertex v and vertex v is called a parent of vertex u if there exists a directed edge from v to u. A vertex is called a leaf if it doesn't have children and has a parent.

Let's call a rooted tree a spruce if its every non-leaf vertex has at least 3 leaf children. You are given a rooted tree, check whether it's a spruce.

The definition of a rooted tree can be found here.

Input

The first line contains one integer n — the number of vertices in the tree (3 ≤ n ≤ 1 000). Each of the next n - 1 lines contains one integer p_i (1 ≤ i ≤ n - 1) — the index of the parent of the i + 1-th vertex (1 ≤ p_i ≤ i).

Vertex 1 is the root. It's guaranteed that the root has at least 2 children.

Output

Print "Yes" if the tree is a spruce and "No" otherwise.

Examples

Input

4
1
1
1

Output

Yes

Input

7
1
1
1
2
2
2

Output

No

Input

8
1
1
1
1
3
3
3

Output

Yes

Note

The first example:

The second example:

It is not a spruce, because the non-leaf vertex 1 has only 2 leaf children.

The third example:

这题就判断了，多开几个数组标记就行，如果不是叶子，那么它就一定要有三个或以上的叶子。

#include <iostream>
using namespace std;
int n;
int k[1010];
int ans[1010];
int vis[1010];
int main(){
    cin>>n;
    for(int i=2;i<=n;i++){
        cin>>k[i];
        ans[k[i]] = 1;
    }
    for(int i=n;i>=2;i--){
        if(vis[i]==0){
            vis[k[i]]++;
        }
    }
    for(int i=1;i<=n;i++){
        if(vis[i]<3&&vis[i]!=0){
            cout<<"No"<<endl;
            return 0;
        }else if(ans[i]&&vis[i]<3){
            cout<<"No"<<endl;
            return 0;
        }
    }
    cout<<"Yes"<<endl;
    return 0;
}

C. Party Lemonade

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

A New Year party is not a New Year party without lemonade! As usual, you are expecting a lot of guests, and buying lemonade has already become a pleasant necessity.

Your favorite store sells lemonade in bottles of n different volumes at different costs. A single bottle of type i has volume 2^i - 1 liters and costs c_i roubles. The number of bottles of each type in the store can be considered infinite.

You want to buy at least L liters of lemonade. How many roubles do you have to spend?

Input

The first line contains two integers n and L (1 ≤ n ≤ 30; 1 ≤ L ≤ 10⁹) — the number of types of bottles in the store and the required amount of lemonade in liters, respectively.

The second line contains n integers c₁, c₂, ..., c_n (1 ≤ c_i ≤ 10⁹) — the costs of bottles of different types.

Output

Output a single integer — the smallest number of roubles you have to pay in order to buy at least L liters of lemonade.

Examples

Input

4 12
20 30 70 90

Output

150

Input

4 3
10000 1000 100 10

Output

10

Input

4 3
10 100 1000 10000

Output

30

Input

5 787787787
123456789 234567890 345678901 456789012 987654321

Output

44981600785557577

Note

In the first example you should buy one 8-liter bottle for 90 roubles and two 2-liter bottles for 30 roubles each. In total you'll get 12 liters of lemonade for just 150 roubles.

In the second example, even though you need only 3 liters, it's cheaper to buy a single 8-liter bottle for 10 roubles.

In the third example it's best to buy three 1-liter bottles for 10 roubles each, getting three liters for 30 roubles.

这题看大佬的代码都是dp,dp,dp, 还有用二进制的，表示不懂，

所以就自己一直找bug,终于写出来了自己的思路的代码。

算是暴力过吧，思想就是找k[i]/list[i]最小的。特别要注意余数。

#include <bits/stdc++.h>
#define ll long long int
#define inf 1000000000000000000
using namespace std;
ll n,m;
ll k[35];
ll List[35];
double vis[35];
double an[35];
int main(){
    cin>>n>>m;
    for(int i=0;i<n;i++){
        cin>>k[i];
    }
    ll sum=1;
    for(int i=0;i<n;i++){
        vis[i] = k[i]*1.0/sum*1.0;
        an[i] = vis[i];
        List[i] = sum;
        sum*=2;
    }
    sort(an,an+n);
    ll cnt=0;
    ll ans = inf;
    ll nt = inf;
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            if(vis[j]==an[i]){
                if(List[j]<=m){
                    ll x = (m/List[j])*k[j];
                    ll y = m/List[j];
                    if(x>=nt*(List[j]*1.0*y/m)){
                        cnt+=nt;
                        m = 0;
                    }else{
                        cnt+=x;
                        m = m%List[j];
                    }
                    ans = min(cnt+k[j],ans),nt = min(k[j],nt);
                    break;
                }
                ans = min(cnt+k[j],ans),nt = min(k[j],nt);
                break;
            }
        }
        if(!m)
            break;
    }
    cnt = min(ans,cnt);
    cout<<cnt<<endl;
    return 0;
}