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 <sys/time.h>
using namespace std;

#define rep(i,n) for(long long i = 0; i < (long long)(n); i++)
#define repi(i,a,b) for(long long i = (long long)(a); i < (long long)(b); i++)
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mt make_tuple
#define mp make_pair
template<class T1, class T2> bool chmin(T1 &a, T2 b) { return b < a && (a = b, true); }
template<class T1, class T2> bool chmax(T1 &a, T2 b) { return a < b && (a = b, true); }

using ll = long long; using vll = vector<ll>; using vvll = vector<vll>; using P = pair<ll, ll>;
ll ugauss(ll a, ll b) { if (!a) return 0; if (a>0^b>0) return a/b; else return (a+(a>0?-1:1))/b+1; }
ll lgauss(ll a, ll b) { if (!a) return 0; if (a>0^b>0) return (a+(a>0?-1:1))/b-1; else return a/b; }
template <typename T, typename U> ostream &operator<<(ostream &o, const pair<T, U> &v) {  o << "(" << v.first << ", " << v.second << ")"; return o; }
template<size_t...> struct seq{}; template<size_t N, size_t... Is> struct gen_seq : gen_seq<N-1, N-1, Is...>{}; template<size_t... Is> struct gen_seq<0, Is...> : seq<Is...>{};
template<class Ch, class Tr, class Tuple, size_t... Is>
void print_tuple(basic_ostream<Ch,Tr>& os, Tuple const& t, seq<Is...>){ using s = int[]; (void)s{0, (void(os << (Is == 0? "" : ", ") << get<Is>(t)), 0)...}; }
template<class Ch, class Tr, class... Args> 
auto operator<<(basic_ostream<Ch, Tr>& os, tuple<Args...> const& t) -> basic_ostream<Ch, Tr>& { os << "("; print_tuple(os, t, gen_seq<sizeof...(Args)>()); return os << ")"; }
ostream &operator<<(ostream &o, const vvll &v) { rep(i, v.size()) { rep(j, v[i].size()) o << v[i][j] << " "; o << endl; } return o; }
template <typename T> ostream &operator<<(ostream &o, const vector<T> &v) { o << '['; rep(i, v.size()) o << v[i] << (i != v.size()-1 ? ", " : ""); o << "]";  return o; }
template <typename T> ostream &operator<<(ostream &o, const deque<T> &v) { o << '['; rep(i, v.size()) o << v[i] << (i != v.size()-1 ? ", " : ""); o << "]";  return o; }
template <typename T>  ostream &operator<<(ostream &o, const set<T> &m) { o << '['; for (auto it = m.begin(); it != m.end(); it++) o << *it << (next(it) != m.end() ? ", " : ""); o << "]";  return o; }
template <typename T>  ostream &operator<<(ostream &o, const unordered_set<T> &m) { o << '['; for (auto it = m.begin(); it != m.end(); it++) o << *it << (next(it) != m.end() ? ", " : ""); o << "]";  return o; }
template <typename T, typename U>  ostream &operator<<(ostream &o, const map<T, U> &m) { o << '['; for (auto it = m.begin(); it != m.end(); it++) o << *it << (next(it) != m.end() ? ", " : ""); o << "]";  return o; }
template <typename T, typename U, typename V>  ostream &operator<<(ostream &o, const unordered_map<T, U, V> &m) { o << '['; for (auto it = m.begin(); it != m.end(); it++) o << *it; o << "]";  return o; }
vector<int> range(const int x, const int y) { vector<int> v(y - x + 1); iota(v.begin(), v.end(), x); return v; }
template <typename T> istream& operator>>(istream& i, vector<T>& o) { rep(j, o.size()) i >> o[j]; return i;}
template <typename T, typename S, typename U> ostream &operator<<(ostream &o, const priority_queue<T, S, U> &v) { auto tmp = v; while (tmp.size()) { auto x = tmp.top(); tmp.pop(); o << x << " ";} return o; }
template <typename T> ostream &operator<<(ostream &o, const queue<T> &v) { auto tmp = v; while (tmp.size()) { auto x = tmp.front(); tmp.pop(); o << x << " ";} return o; }
template <typename T> ostream &operator<<(ostream &o, const stack<T> &v) { auto tmp = v; while (tmp.size()) { auto x = tmp.top(); tmp.pop(); o << x << " ";} return o; }
template <typename T> unordered_map<T, ll> counter(vector<T> vec){unordered_map<T, ll> ret; for (auto&& x : vec) ret[x]++; return ret;};
void vizGraph(vvll& g, int mode = 0, string filename = "out.png") { ofstream ofs("./out.dot"); ofs << "digraph graph_name {" << endl; set<P> memo; rep(i, g.size())  rep(j, g[i].size()) { if (mode && (memo.count(P(i, g[i][j])) || memo.count(P(g[i][j], i)))) continue; memo.insert(P(i, g[i][j])); ofs << "    " << i << " -> " << g[i][j] << (mode ? " [arrowhead = none]" : "")<< endl;  } ofs << "}" << endl; ofs.close(); system(((string)"dot -T png out.dot >" + filename).c_str()); }
struct timeval start; double sec() { struct timeval tv; gettimeofday(&tv, NULL); return (tv.tv_sec - start.tv_sec) + (tv.tv_usec - start.tv_usec) * 1e-6; }
size_t random_seed; struct init_{init_(){ ios::sync_with_stdio(false); cin.tie(0); gettimeofday(&start, NULL); struct timeval myTime; struct tm *time_st; gettimeofday(&myTime, NULL); time_st = localtime(&myTime.tv_sec); srand(myTime.tv_usec); random_seed = RAND_MAX / 2 + rand() / 2; }} init__;
#define ldout fixed << setprecision(40) 

#define EPS (double)1e-14
#define INF (ll)1e18
#define mo  (ll)(1e9+7)

int main(void) {
    double x, p; cin >> x >> p; p /= 100;
    double ret = 0;
    if ((ll)(x) % 2 == 0) {
        ret = x / 2. / p;
    } else {
        ret = (x-1.)/2.+(1.-p)*(x+1.)/2./p+1;
    }

    cout << ldout << ret << endl;

    return 0;
}