Problem:
Score : 700 points

Problem Statement

Aoki is in search of Takahashi, who is missing in a one-dimentional world.
Initially, the coordinate of Aoki is 0, and the coordinate of Takahashi is known to be x, but his coordinate afterwards cannot be known to Aoki.

Time is divided into turns. In each turn, Aoki and Takahashi take the following actions simultaneously:

Let the current coordinate of Aoki be a, then Aoki moves to a coordinate he selects from a-1, a and a+1.

Let the current coordinate of Takahashi be b, then Takahashi moves to the coordinate b-1 with probability of p percent, and moves to the coordinate b+1 with probability of 100-p percent.

When the coordinates of Aoki and Takahashi coincide, Aoki can find Takahashi.
When they pass by each other, Aoki cannot find Takahashi.

Aoki wants to minimize the expected number of turns taken until he finds Takahashi. Find the minimum possible expected number of turns.

Constraints

1 ≦ x ≦ 1,000,000,000

1 ≦ p ≦ 100

x and p are integers.

Partial Scores

In the test set worth 200 points, p=100.

In the test set worth 300 points, x ≦ 10.

Input

The input is given from Standard Input in the following format:

x
p

Output

Print the minimum possible expected number of turns.
The output is considered correct if the absolute or relative error is at most 10^{-6}.

Sample Input 1

3
100

Sample Output 1

2.0000000

Takahashi always moves by -1.
Thus, by moving to the coordinate 1 in the 1-st turn and staying at that position in the 2-nd turn, Aoki can find Takahashi in 2 turns.

Sample Input 2

6
40

Sample Output 2

7.5000000

Sample Input 3

101
80

Sample Output 3

63.7500000

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef signed long long ll;

#undef _P
#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
//-------------------------------------------------------

ll X;
double P;

void solve() {
	int i,j,k,l,r,x,y; string s;
	
	cin>>X>>P;
	
	if(P!=100) return;
	P/=100;
	_P("%lld\n",(X+1)/2);
	
}


int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n';
	FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	solve(); return 0;
}