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 <bits/stdc++.h>
#include <random>

using namespace std;

typedef unsigned long long ull;
typedef long long ll;
typedef double ld;
//#define ll __int128
//#define int ll
//#define char ll
//#define double ld
typedef vector<char> vc;
typedef vector <vc> vvc;
typedef vector <vvc> vvvc;
typedef pair<int, int> pii;
typedef pair<pii, pii> piii;
typedef pair<ll, ll> pll;
typedef vector <int> vi;
typedef vector <pii> vpi;
typedef vector <vi> vvi;
typedef vector <vvi> vvvi;
typedef vector<short> vs;
typedef vector <vs> vvs;
typedef vector <vvs> vvvs;
typedef vector <ll> vl;
typedef vector <vl> vvl;
typedef vector <vvl> vvvl;
typedef vector <ld> vld;
typedef vector <vld> vvld;
typedef vector <vvld> vvvld;
typedef vector <string> vst;
typedef vector <vst> vvst;
typedef pair<ld, ld> pld;

#define inmin(a, b) a = min(a, (b))
#define inmax(a, b) a = max(a, (b))
#define ALL(a) a.begin(),a.end()
#define RALL(a) a.rbegin(),a.rend()
#define sqr(x) ((x) * (x))
#define fori(i, n) for(int i = 0; i < int(n); ++i)
#define SZ(a) ((int)((a).size()))
#define triple(T) tuple<T, T, T>
#define quad(T) tuple<T, T, T, T>
#define watch(x) cerr << (#x) << " = " << (x) << endl;

#ifdef MAX_HOME
#define cerr cout
#else
#define cerr if (false) cerr
#endif

const double PI = 2 * acos(0.0);
#define rand shittttty_shit
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng_64(chrono::steady_clock::now().time_since_epoch().count());

const string DIGITS = "0123456789";
const string ALPH = "abcdefghijklmnopqrstuvwxyz";


template<class T0, class T1>
inline ostream &operator<<(ostream &out, pair<T0, T1> &a) {
    return out << "{" << a.first << ", " << a.second << "}";
}

template<class T0, class T1>
inline istream &operator>>(istream &in, pair<T0, T1> &a) {
    return in >> a.first >> a.second;
}

template<class T0, class T1, class T2>
inline ostream &operator<<(ostream &out, tuple<T0, T1, T2> &a) {
    return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a) << "}";
}

template<class T0, class T1, class T2, class T3>
inline ostream &operator<<(ostream &out, tuple<T0, T1, T2, T3> &a) {
    return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a) << ", " << get<3>(a) << "}";
}

template<class T>
inline ostream &operator<<(ostream &out, vector <T> &a) {
    out << "[";
    fori (i, a.size())out << a[i] << vector < string > {", ", "]  "}[i + 1 == a.size()];
    return out;
}


void smain();


signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
#ifdef MAX_HOME
    freopen("input.txt", "r", stdin);
    clock_t start = clock();
#endif
#ifdef HK_CUP
    freopen("finput.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif

    cout << setprecision(12) << fixed;
    smain();
#ifdef MAX_HOME
#ifndef HK_CUP
    cout << "\n\nTOTAL EXECUTION TIME: " << float(clock() - start) / CLOCKS_PER_SEC << endl;
#endif
#endif
    return 0;
}

ld f(ll x, ld p) {
    return x / p;
}

void smain() {
    ll x;
    ld p;
    cin >> x >> p;
    p /= 100;
    ld ans = 0;
    if (x & 1) {
        ans = 1 + p * f((x - 1) / 2, p) + (1 - p) * f((x + 1) / 2, p);
    } else {
        ans = f(x / 2, p);
    }
    cout << ans;
}