Problem statement

Real variables $ x_1, x_2, ..., x_N $ satisfy the following conditions.

1. $ 0 \ leq x_i \ leq 1 $ ($ 1 \ leq i \ leq N $)
2. $ w_1x_1 + w_2x_2 + ... + w_Nx_N \ leq W $



At this time, find the maximum value that $ v_1x_1 + v_2x_2 + ... + v_Nx_N $ can take. It is known that such a maximum actually exists.

Constraint

* $ 1 \ leq N \ leq 10 ^ 5 $
* $ 1 \ leq W \ leq 10 ^ 5 $
* $ -10 ^ 4 \ leq w_i \ leq 10 ^ 4 $
* $ -10 ^ 4 \ leq v_i \ leq 10 ^ 4 $



input

Input follows the following format. All given numbers are integers.


$ N $ $ W $
$ w_1 $ $ v_1 $
$ w_2 $ $ v_2 $
$ ... $
$ w_N $ $ v_N $

output

Output the maximum possible value of $ v_1x_1 + v_2x_2 + ... + v_Nx_N $ on one line. The output must not have an error greater than $ 10 ^ {-3} $.

Examples

Input

1 1
3 1


Output

0.333333


Input

2 3
3 3
1 2


Output

4.000000


Input

2 1
-1 -3
3 10


Output

3.666667
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstdio>

#define EPS 1e-9

using namespace std;

int n;
double W;

bool pred(pair<int,int> a, pair<int,int> b){
    return a.first * b.second > a.second * b.first;
}

int main(){
    cin >> n >> W;
    vector<pair<int,int> > vp;
    double ret=0;
    double w,v;
    for(int i=0;i<n;i++){
        cin >> w >> v;
        if(w <= 0 && v >= 0){
            W -= w;
            ret += v;
        }else if(w >=0 && v <= 0){
            //do nothing
        }else if(v < 0 && w < 0){
            W -= w;
            ret += v;
            vp.push_back(make_pair(-v,-w));
        }else{
            vp.push_back(make_pair(v,w));
        }
    }
    sort(vp.begin(), vp.end(), pred);
    for(int i=0;i<vp.size();i++){
        double tmp = min(1.0, W/vp[i].second);
        ret += vp[i].first * tmp;
        W -= tmp * vp[i].second;
        if(W < EPS) break;
    }
    printf("%.9f\n", ret);
}