Problem: 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.

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}.

Examples

Input

3
100


Output

2.0000000


Input

6
40


Output

7.5000000


Input

101
80


Output

63.7500000

Solution (in CPP):

Solution:
#include <iostream>
#include <cstdio>
#include <vector>
#include <cmath>
#include <map>
#include <algorithm>
#include <string>
#include <utility>
#include <set>
#include <stack>
#include <deque>
#include <ctime>
#include <random>
#include <cassert>
#include <cmath>
#include <climits>
#include <queue>
#include <cstring>
#include <bitset>
#include <iomanip>
#include <chrono>

/*
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("popcnt")
#pragma GCC target("avx2")
*/


#ifdef LOCAL
    #define dbg(x) cout << #x << " : " << x << endl;
#else 
    #define dbg(x)
#endif

#define int long long
#define pb push_back
#define ppb pop_back()
#define mp make_pair
#define fi(a,b) for (int i=a;i<b;i++)
#define fj(a,b) for (int j=a;j<b;j++)
#define fk(a,b) for (int k=a;k<b;k++)
#define fi1(a,b) for (int i=a-1;i>=b;i--)
#define fj1(a,b) for (int j=a-1;j>=b;j--)
#define fk1(a,b) for (int k=a-1;k>=b;k--)
#define fx(x,a) for (auto& x : a)
#define rep(i, a, b) for (int i = a; i < b; ++i)
#define rep1(i, a, b) for (int i = a - 1; i >= b; --i)
#define siz(x) (int)x.size()
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()

using namespace std;

template<typename T1, typename T2>inline void mine(T1 &x, T2 y) { if (x > y) x = y; }
template<typename T1, typename T2>inline void maxe(T1 &x, T2 y) { if (x < y) x = y; }

ostream& operator << (ostream &a, const vector<int> &b) {
    for (auto k : b) cout << k << ' ';
    return a;
}


typedef long long ll;
typedef unsigned long long ull;
typedef char ch;
typedef string str;
typedef pair<int, int> pii;
typedef map<int, int> mii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vpii;
typedef vector<vpii> vvpii;
typedef vector<ch> vch;
typedef vector<vch> vvch;
typedef vector<str> vs;



const int MOD = 1000000007;
const int INF = 1000000050;
const long long BIG = (long long)2e18 + 50;
const int MX = 200010;
const double EPS = 1e-9;


mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

double p;

double M(int x) {
    return x / (2 * p);
}

int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    
    int x;
    cin >> x >> p;
    p /= 100;
    double ans;
    if (x % 2)
        ans = M(x - 1) * p + M(x + 1) * (1 - p) + 1;
    else
        ans = M(x);
    cout << setprecision(15) << fixed;
    cout << ans << '\n';     
}