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;
int Day12[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int prime100[] = {2,  3,  5,  7,  11, 13, 17, 19, 23, 29, 31, 37,  41, 43,
                  47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103};
template <typename T>
inline bool isLeap(T y) {
  return (y % 400 == 0) || (y % 100 ? y % 4 == 0 : false);
}
template <typename T>
inline T GCD(T a, T b) {
  a = abs(a);
  b = abs(b);
  if (a < b) swap(a, b);
  while (b) {
    a = a % b;
    swap(a, b);
  }
  return a;
}
template <typename T>
inline T LCM(T x, T y) {
  T tp = GCD(x, y);
  if ((x / tp) * 1. * y > 9e18) return 9e18;
  return (x / tp) * y;
}
template <typename T>
inline T BIGMOD(T A, T B, T M = 1000000007) {
  T ret = 1;
  while (B) {
    if (B & 1) ret = (ret * A) % M;
    A = (A * A) % M;
    B = B >> 1;
  }
  return ret;
}
template <typename T>
inline T BigMod(T A, T B, T M) {
  T ret = 1;
  while (B) {
    if (B & 1) ret = (ret * A) % M;
    A = (A * A) % M;
    B = B >> 1;
  }
  return ret;
}
long long int MySqrt(long long int n) {
  long long int p = sqrt(n);
  if ((p + 1) * (p + 1) <= n)
    return p + 1;
  else if (p * p <= n)
    return p;
  else
    return p - 1;
}
long long int MyPow(long long int x, long long int n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return MyPow(x * x, n / 2);
  else
    return x * MyPow(x * x, ((n - 1) / 2));
}
long long int modInverse(long long int n) {
  return BIGMOD(n, (long long int)1000000007 - 2) % 1000000007;
}
int main() {
  double H, L;
  cin >> H >> L;
  double Low = 0, High = L * L;
  while (Low < High) {
    double mid = (Low + High) / 2.0;
    double a = H * H + 2.0 * H * mid;
    double b = L * L;
    if (abs(a - b) < 0.0000001) {
      printf("%0.8f\n", mid);
      return 0;
    }
    if (a > b)
      High = mid;
    else
      Low = mid;
  }
}