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>

using namespace std;

typedef long long ll;
typedef complex<double> point;
#define mapii map<int, int>
#define debug(a) cout << #a << ": " << a << endl
#define debuga1(a, l, r) fto(i, l, r) cout << a[i] << " "; cout << endl
#define fdto(i, r, l) for(int i = (r); i >= (l); --i)
#define fto(i, l, r) for(int i = (l); i <= (r); ++i)
#define forit(it, var) for(__typeof(var.begin()) it = var.begin(); it != var.end(); it++)
#define forrit(rit, var) for(__typeof(var.rbegin()) rit = var.rbegin(); rit != var.rend(); rit++)
#define ii pair<int, int>
#define iii pair<int, ii>
#define ff first
#define ss second
#define mp make_pair
#define pb push_back
#define maxN 100005
#define MOD 1000
#define oo 1000000000000000007LL
#define sz(a) (int)a.size()

const double PI = acos(-1.0);

double fRand(double fMin, double fMax)
{
    double f = (double)rand() / RAND_MAX;
    return fMin + f * (fMax - fMin);
}

template <class T>
T min(T a, T b, T c) {
    return min(a, min(b, c));
}

template <class T>
T max(T a, T b, T c) {
    return max(a, max(b, c));
}

int x, p;

int main () {
    scanf("%d%d", &x, &p);

    double ans = (x+1)/2 * 100.0/p;
    printf("%.9g", ans);

    return 0;
}