The UNAL programming coaches have lost a bet, they bet the 6 UNAL teams would occupy the first six positions in the regional finals. Now, they have to shave their heads!. Some of them have more hair than others, and consequently, it takes more time to shave a head completely. However, all of the coaches really love their hair, therefore there is a probability that some (posibly all) of them daunt at the very last moment and do not permit the hairdresser to shave their heads.

Norbert, the hairdresser who loves probability, would like to order the coaches' schedule such that the average time in the hair salon is minimized for all the coaches. First all the coaches are there at the same time, then they start going one by one to Norbert, if by the moment a coach has to go to the hairdresser, he or she daunts then he or she simply leaves the hair salon and it is the turn of the next coach, after the head of a coach is shaved then that coach leaves the hair salon. The time between turns is negligible.

For example, suppose that shaving Diego's head takes 2 minutes and shaving Ivan's takes 3 minutes, but Diego has probability of 0.5 of not daunting meanwhile Ivan for sure will shave his head. If Ivan goes first, he will stay 3 minutes in the hair salon and Diego will stay there 3 minutes if daunting or 5 minutes if not (3 of them waiting for Ivan to finish), in this case the average expected time of the coaches in the hair salon would be 3.5, note this is not the optimal arrangement.

Now, Norbert knows the time it takes to shave each head and the probability of a coach to accept to have the head shaved in the barbershop, help him to know the minimum average expected time in the hair salon of the coaches.

Input

The first line of input is an integer n (1 ≤ n ≤ 5 * 10⁵) - the number of coaches.

The next n lines contain each an integer x (0 ≤ x ≤ 100) and a decimal number y (0 ≤ y ≤ 1) separated by a single space - the time in minutes it takes to shave the head of the i - th coach and his probability of not daunting, respectively.

Output

Print the minimum expected average time. The relative error of your answer should not exceed 10^- 6.

Examples

Input

2 2 0.5 3 1.0

Output

2.500000000

Input

2 0 0.4 20 0.6

Output

6.000000000 从小到大排序，然后累计即可；

#include
        
         using namespace std;#define maxn 600005#define inf 999999999999//#define INF 1e18#define rdint(x) scanf("%d",&x)#define rdllt(x) scanf("%lld",&x)#define rdult(x) scanf("%lu",&x)#define rdlf(x) scanf("%lf",&x)#define rdstr(x) scanf("%s",x)typedef long long  ll;const long long int mod=1e9+7;#define ms(x) memset((x),0,sizeof(x))int n;struct node{	int x;	double prop;}t[maxn];double ans=0.0;double tmp[maxn];main(){ //   ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);    cin>>n;    for(int i=1;i<=n;i++)rdint(t[i].x),rdlf(t[i].prop),tmp[i]=1.0*t[i].x*t[i].prop;    sort(tmp+1,tmp+1+n);    for(int i=1;i<=n;i++){    	ans+=tmp[i]*1.0*(n-i+1);	}	printf("%.7lf\n",1.0*ans/(double)n);	}