Consider a system of n water taps all pouring water into the same container. The i-th water tap can be set to deliver any amount of water from 0 to ai ml per second (this amount may be a real number). The water delivered by i-th tap has temperature ti.

If for every <image> you set i-th tap to deliver exactly xi ml of water per second, then the resulting temperature of water will be <image> (if <image>, then to avoid division by zero we state that the resulting water temperature is 0).

You have to set all the water taps in such a way that the resulting temperature is exactly T. What is the maximum amount of water you may get per second if its temperature has to be T?

Input

The first line contains two integers n and T (1 ≤ n ≤ 200000, 1 ≤ T ≤ 106) — the number of water taps and the desired temperature of water, respectively.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 106) where ai is the maximum amount of water i-th tap can deliver per second.

The third line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 106) — the temperature of water each tap delivers.

Output

Print the maximum possible amount of water with temperature exactly T you can get per second (if it is impossible to obtain water with such temperature, then the answer is considered to be 0).

Your answer is considered correct if its absolute or relative error doesn't exceed 10 - 6.

Examples

Input

2 100
3 10
50 150


Output

6.000000000000000


Input

3 9
5 5 30
6 6 10


Output

40.000000000000000


Input

2 12
1 3
10 15


Output

1.666666666666667
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
const double eps = 1e-16;
const int inf = 2000000000;
long long int MOD = 1000000007;
int MOD1 = 1000000007;
int MOD2 = 1000000009;
inline bool checkBit(long long int n, long long int i) {
  return n & (1LL << i);
}
inline long long int setBit(long long int n, long long int i) {
  return n | (1LL << i);
  ;
}
inline long long int resetBit(long long int n, long long int i) {
  return n & (~(1LL << i));
}
int dx[] = {0, 0, +1, -1};
int dy[] = {+1, -1, 0, 0};
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline bool isLeapYear(long long int year) {
  return (year % 400 == 0) || (year % 4 == 0 && year % 100 != 0);
}
inline void normal(long long int &a) {
  a %= MOD;
  (a < 0) && (a += MOD);
}
inline long long int modMul(long long int a, long long int b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  return (a * b) % MOD;
}
inline long long int modAdd(long long int a, long long int b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  return (a + b) % MOD;
}
inline long long int modSub(long long int a, long long int b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long int modPow(long long int b, long long int p) {
  long long int r = 1LL;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1LL;
  }
  return r;
}
inline long long int modDiv(long long int a, long long int b) {
  return modMul(a, modPow(b, MOD - 2));
}
bool comp(const pair<long long int, pair<long long int, long long int> > &p1,
          const pair<long long int, pair<long long int, long long int> > &p2) {
  return p1.first > p2.first;
}
bool comp1(const pair<long long int, long long int> &p1,
           const pair<long long int, long long int> &p2) {
  if (p1.first == p2.first) {
    return p1.second > p2.second;
  }
  return p1.first < p2.first;
}
long long int converter(string a) {
  long long int i, mul = 1, r, t, ans = 0LL;
  if (a.length() == 0) return 0;
  for (i = a.length() - 1; i >= 0; i--) {
    t = a[i] - '0';
    r = t % 10;
    ans += (mul * r);
    mul = mul * 10;
  }
  return ans;
}
int msb(unsigned x) {
  union {
    double a;
    int b[2];
  };
  a = x;
  return (b[1] >> 20) - 1023;
}
const int MAX = 200005;
double T;
pair<double, double> p[MAX];
int n;
bool func(double x) {
  double need = x * T;
  double have = 0.0;
  for (int i = 1; i <= n; ++i) {
    if ((x - p[i].second) >= eps) {
      double cur = (p[i].second * p[i].first);
      have += cur;
      x -= p[i].second;
    } else {
      double cur = p[i].first * x;
      have += cur;
      x = 0.0;
    }
  }
  double have2 = 0.0;
  x = need / T;
  for (int i = n; i >= 1; --i) {
    if ((x - p[i].second) >= eps) {
      double cur = (p[i].second * p[i].first);
      have2 += cur;
      x -= p[i].second;
    } else {
      double cur = p[i].first * x;
      have2 += cur;
      x = 0.0;
    }
  }
  return (need >= have && need <= have2);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cout.unsetf(ios::floatfield);
  cout.precision(14);
  cout.setf(ios::fixed, ios::floatfield);
  ;
  int i;
  cin >> n >> T;
  for (i = 1; i <= n; ++i) cin >> p[i].second;
  for (i = 1; i <= n; ++i) cin >> p[i].first;
  sort(p + 1, p + n + 1);
  double low = 0.0, high = 9000000000000.0, mid, ans = 0.0;
  for (i = 1; i <= 150; ++i) {
    mid = ((low + high) * 1.0) / 2;
    if (func(mid) == true) {
      ans = max(mid, ans);
      low = mid;
    } else
      high = mid;
  }
  cout << ans << '\n';
  return 0;
}
