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 <iomanip>
#include <algorithm>
#include <bitset>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>
#include <cmath>
#include <time.h>
#include <random>
#include <string>
#include <cassert>
#include <vector>
#include <ostream>
#include <istream>
#include <stack>
#include <deque>
#include <queue>
#include <functional>
#include <chrono>
#include <stack>

using namespace std;

#define int long long
#define pb push_back
#define all(a) (a).begin(), (a).end()
#define pii pair<int, int>
#define ld long double

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

istream& operator>> (istream& in, pii& b) {
    in >> b.first >> b.second;
    return in;
}

ostream& operator<< (ostream& out, const pii& b) {
    out << "{" << b.first << ", " << b.second << "}";
    return out;
}

template <typename T1, typename T2> inline bool chkmin(T1 &x, const T2 &y) {if (x > y) {x = y; return 1;} return 0;}
template <typename T1, typename T2> inline bool chkmax(T1 &x, const T2 &y) {if (x < y) {x = y; return 1;} return 0;}


#ifdef LOCAL
    #define dbg(x) cout << #x << " : " << (x) << "\n";
    const int INF = 1e18;
    // const int mod = 2600000069;
    // const int p = 10;
#else
    #define dbg(x) 57
    const int INF = 1e18;
    // const int mod = 2600000069;
    // const int p = 179;
#endif

const ld PI = acos(-1);

#define time clock() / (double) CLOCKS_PER_SEC

// #pragma GCC optimize("Ofast,no-stack-protector")
// #pragma GCC target("sse,sse2,sse3,sse3,sse4")
// #pragma GCC optimize("unroll-loops")
// #pragma GCC optimize("fast-math")
// #pragma GCC target("avx2")  
// #pragma GCC optimize("section-anchors")
// #pragma GCC optimize("profile-values,profile-reorder-functions,tracer")
// #pragma GCC optimize("vpt")
// #pragma GCC optimize("rename-registers")
// #pragma GCC optimize("move-loop-invariants")
// #pragma GCC optimize("unswitch-loops")
// #pragma GCC optimize("function-sections")
// #pragma GCC optimize("data-sections")

mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());

signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int x, p;
    cin >> x >> p;
    cout << fixed << setprecision(8);
    if (x & 1) {
        cout << ((100.0 - p) / 100.0 * (ld)(x + 1) / (ld)p * 100.0 + (ld)p / 100.0 * (ld)(x - 1) / (ld)p * 100.0) / 2.0 + 1.0 << "\n";
    } else {
        cout << ((ld)x / (ld)p * 100.0) / 2.0 << "\n";
    }
}
/*

*/