Codeforces 801C(二分)

问题描述:

You have n devices that you want to use simultaneously.

The i-th device uses ai units of power per second. This usage is continuous. That is, in λ seconds, the device will use λ·ai units of power. The i-th device currently has bi units of power stored. All devices can store an arbitrary amount of power.

You have a single charger that can plug to any single device. The charger will add punits of power per second to a device. This charging is continuous. That is, if you plug in a device for λ seconds, it will gain λ·p units of power. You can switch which device is charging at any arbitrary unit of time (including real numbers), and the time it takes to switch is negligible.

You are wondering, what is the maximum amount of time you can use the devices until one of them hits 0 units of power.

If you can use the devices indefinitely, print -1. Otherwise, print the maximum amount of time before any one device hits 0 power.

Input

The first line contains two integers, n and p (1 ≤ n ≤ 100 000, 1 ≤ p ≤ 109) — the number of devices and the power of the charger.

This is followed by n lines which contain two integers each. Line i contains the integers ai and bi (1 ≤ ai, bi ≤ 100 000) — the power of the device and the amount of power stored in the device in the beginning.

Output

If you can use the devices indefinitely, print -1. Otherwise, print the maximum amount of time before any one device hits 0 power.

Your answer will be considered correct if its absolute or relative error does not exceed 10 - 4.

Namely, let's assume that your answer is a and the answer of the jury is b. The checker program will consider your answer correct if .

Example

Input

2 1
2 2
2 1000

Output

2.0000000000

Input

1 100
1 1

Output

-1

Input

3 5
4 3
5 2
6 1

Output

0.5000000000

题目题意:N个机器，每个机器每秒钟需要消耗能量，每台机器原本储存着一些能量，我们有一个充电器每秒钟可以提供一些能量，问最长可以坚持多少时间（所有机器都有能量）。

题目分析:二分时间，我们只需保证在这个时间内，机器消耗的能量充电器可以提供，那么就可以。二分的时候时间的上限注意开大一点，错了好久的，精度不必太高不然会T

代码如下:

#include
#include
#include
#include
using namespace std;

const int maxn=1e5+100;
const double epx=1e-5;
int a[maxn],b[maxn],n;
double p,suma;

bool check(double t)
{
     double sum=0;
     for (int i=1;i<=n;i++) {
        if (a[i]*t-epx) return false;
     else return true;
}
int main()
{
    scanf("%d%lf",&n,&p);
    for (int i=1;i<=n;i++) {
        scanf("%d%d",&a[i],&b[i]);
        suma+=a[i];
    }
    if (p-suma>-epx) {
        printf("-1\n");
        return 0;
    }
    double left=0,right=1e15,mid;
    while (right-left>epx) {
        mid=(left+right)/2;
        if (check(mid)) left=mid;
        else right=mid;
    }
    printf("%.6lf\n",left);
    return 0;
}