Problem: While sailing on a boat, Inessa noticed a beautiful water lily flower above the lake's surface. She came closer and it turned out that the lily was exactly H centimeters above the water surface. Inessa grabbed the flower and sailed the distance of L centimeters. Exactly at this point the flower touched the water surface.

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Suppose that the lily grows at some point A on the lake bottom, and its stem is always a straight segment with one endpoint at point A. Also suppose that initially the flower was exactly above the point A, i.e. its stem was vertical. Can you determine the depth of the lake at point A?

Input

The only line contains two integers H and L (1 ≤ H < L ≤ 10^{6}).

Output

Print a single number — the depth of the lake at point A. The absolute or relative error should not exceed 10^{-6}.

Formally, let your answer be A, and the jury's answer be B. Your answer is accepted if and only if \frac{|A - B|}{max{(1, |B|)}} ≤ 10^{-6}.

Examples

Input


1 2


Output


1.5000000000000


Input


3 5


Output


2.6666666666667

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
ifstream fin("input.txt");
ofstream fout("output.txt");
bool isPrime(long long int x) {
  if (x <= 4 || x % 2 == 0 || x % 3 == 0) return x == 2 || x == 3;
  for (int i = 5; i * i <= x; i += 6)
    if (x % i == 0 || x % (i + 2) == 0) return 0;
  return 1;
}
void primeFactors(long long n, set<long long> &Fac) {
  while (n % 2 == 0) {
    Fac.insert(2);
    n = n / 2;
  }
  for (int i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      Fac.insert(i);
      n = n / i;
    }
  }
  if (n > 2) Fac.insert(n);
}
int NumDigits(int x) {
  return (
      x < 10
          ? 1
          : (x < 100
                 ? 2
                 : (x < 1000
                        ? 3
                        : (x < 10000
                               ? 4
                               : (x < 100000
                                      ? 5
                                      : (x < 1000000
                                             ? 6
                                             : (x < 10000000
                                                    ? 7
                                                    : (x < 100000000
                                                           ? 8
                                                           : (x < 1000000000
                                                                  ? 9
                                                                  : 10)))))))));
}
inline void boostIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout.precision(10);
}
long long int Pow10[19] = {
    (long long int)0,    (long long int)1e1,  (long long int)1e2,
    (long long int)1e3,  (long long int)1e4,  (long long int)1e5,
    (long long int)1e6,  (long long int)1e7,  (long long int)1e8,
    (long long int)1e9,  (long long int)1e10, (long long int)1e11,
    (long long int)1e12, (long long int)1e13, (long long int)1e14,
    (long long int)1e15, (long long int)1e16, (long long int)1e17,
    (long long int)1e18};
long long int gcd(long long int a, long long int b) {
  return b ? gcd(b, a % b) : a;
}
long long int lcm(long long int a, long long int b) {
  return a * b / gcd(a, b);
}
long long MOD = 1e9 + 7;
template <class T>
struct Fenwick {
  vector<T> V;
  Fenwick(long long n) : V(n) {}
  void add(size_t i, T x) {
    for (; i < V.size(); i |= i + 1) V[i] += x;
  }
  T sum(long long i) {
    T r = T();
    for (--i; i >= 0; i = (i & (i + 1)) - 1) r += V[i];
    return r;
  }
  T sum(long long l, long long r) { return sum(r) - sum(l); }
};
long long f_Minimum(long long a, long long b) { return (min(a, b)); }
long long f_Maximum(long long a, long long b) { return (max(a, b)); }
long long f_Summarize(long long a, long long b) { return (a + b); }
class Parameter {
 public:
  long long InitialValue;
  long long (*Function)(long long a, long long b);
  Parameter(long long (*f)(long long a, long long b), long long value) {
    Function = f;
    InitialValue = value;
  }

 private:
};
Parameter Minimum(f_Minimum, 9223372036854775807);
Parameter Maximum(f_Maximum, -9223372036854775807);
Parameter Summarize(f_Summarize, 0);
class SegmentTree {
 public:
  SegmentTree(long long size, Parameter p) {
    Function = p.Function;
    InitialValue = p.InitialValue;
    n = size;
    data = vector<long long>(2 * n, 0);
  }
  SegmentTree(vector<long long> array, Parameter p) {
    Function = p.Function;
    InitialValue = p.InitialValue;
    n = array.size();
    ;
    data = vector<long long>(2 * n, 0);
    for (long long i = 0; i < (n); ++i) {
      data[n + i] = array[i];
    }
    for (long long i = n - 1; i >= 1; ++i) {
      data[i] = Function(data[i * 2], data[i * 2] + 1);
    }
  }
  void Update(long long idx, long long value) {
    idx += n;
    data[idx] = value;
    while (idx > 1) {
      idx /= 2;
      data[idx] = Function(data[idx * 2], data[idx * 2 + 1]);
    }
  }
  long long Find(long long left, long long right) {
    left += n;
    right += n;
    long long value = InitialValue;
    while (left < right) {
      if (left % 2 == 1) {
        value = Function(left, value);
        left++;
      }
      if (right % 2 == 1) {
        value = Function(right, value);
        right--;
      }
      left /= 2;
      right /= 2;
    }
    return value;
  }

 private:
  long long n;
  vector<long long> data;
  long long (*Function)(long long a, long long b);
  long long InitialValue;
};
int32_t main() {
  boostIO();
  long double H, L;
  cin >> H >> L;
  long double z = sqrt(H * H + L * L);
  cerr << "z = " << z << '\n';
  long double cos = H / z;
  cerr << "cos = " << cos << '\n';
  long double k = z / 2;
  long double A = k / cos;
  cout << A - H << '\n';
  cerr << '\n' << "RUNTIME: " << 1.0 * clock() / CLOCKS_PER_SEC << " s" << '\n';
  return 0;
  ;
}