Codeforces 801C Voltage Keepsake【二分】

C. Voltage Keepsake

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You have n devices that you want to use simultaneously.

The i-th device uses a_i units of power per second. This usage is continuous. That is, in λ seconds, the device will use λ·a_i units of power. The i-th device currently has b_i 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 p units 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 a_i and b_i (1 ≤ a_i, b_i ≤ 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 .

Examples

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

Note

In sample test 1, you can charge the first device for the entire time until it hits zero power. The second device has enough power to last this time without being charged.

In sample test 2, you can use the device indefinitely.

In sample test 3, we can charge the third device for 2 / 5 of a second, then switch to charge the second device for a 1 / 10 of a second.

题目大意：

给你N个需要充电的物品，每个物品一开始有bi的电量，每一单位时间消耗ai的电量。现在有一个充电器，每一单位时间可以补充P点电量，问你第一次出现有物品没电了的时刻。

思路：

时间具有单调性，那么考虑二分时间，接下来对于每个物品的时间总用电量进行统计，然后对应将每个物品需要的充电量进行统计，对于这个统计的量和p比较大小即可。

如果这个统计量小于等于p，那么增大时间，相反，减小时间。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<math.h>
#include<algorithm>
using namespace std;
#define eps 1e-6
struct node
{
    double x,y;
}a[108000];
int n;
double p;
int Slove(double mid)
{
    double sum=0;
    for(int i=0;i<n;i++)
    {
        double yu=a[i].y-a[i].x*mid;
        if(yu<0)sum+=fabs(yu);
    }
    if(sum+eps<=p*mid)
    {
        return 1;
    }
    else return 0;
}
int main()
{
    while(~scanf("%d%lf",&n,&p))
    {
        for(int i=0;i<n;i++)scanf("%lf%lf",&a[i].x,&a[i].y);

        double ans=-1;
        double l=0;
        double r=120;
        if(Slove(r)==0)
        {
            printf("-1\n");
            continue;
        }
        for(int i=0;i<80;i++)
        {
            double mid=(l+r)/2;
            if(Slove(mid)==1)
            {
                ans=mid;
                l=mid;
            }
            else r=mid;
        }
        if(ans>1200000000000)printf("-1\n");
        else
        printf("%lf\n",ans);
    }
}