Problem:
Score : 300 points

Problem Statement

Hearing that energy drinks increase rating in those sites, Takahashi decides to buy up M cans of energy drinks.

There are N stores that sell energy drinks. In the i-th store, he can buy at most B_i cans of energy drinks for A_i yen (the currency of Japan) each.

What is the minimum amount of money with which he can buy M cans of energy drinks?

It is guaranteed that, in the given inputs, a sufficient amount of money can always buy M cans of energy drinks.

Constraints

All values in input are integers.

1 \leq N, M \leq 10^5

1 \leq A_i \leq 10^9

1 \leq B_i \leq 10^5

B_1 + ... + B_N \geq M

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
A_2 B_2
\vdots
A_N B_N

Output

Print the minimum amount of money with which Takahashi can buy M cans of energy drinks.

Sample Input 1

2 5
4 9
2 4

Sample Output 1

12

With 12 yen, we can buy one drink at the first store and four drinks at the second store, for the total of five drinks. However, we cannot buy 5 drinks with 11 yen or less.

Sample Input 2

4 30
6 18
2 5
3 10
7 9

Sample Output 2

130

Sample Input 3

1 100000
1000000000 100000

Sample Output 3

100000000000000

The output may not fit into a 32-bit integer type.

Solution in C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>

int main(){
    int N,M;
    double A,B;
    double costs=0;
    int i,j,k;

    scanf("%d %d",&N,&M);
    double limit[N],price[N];

    for(i=0;i<N;i++){
        scanf("%lf %lf",&A,&B);
        if(i==0){
            price[0]=A;
            limit[0]=B;
        }else{
            for(j=0;j<i+1;j++){
                if(A<=price[j]){
                    for(k=N-1;k>=j-1;k--){
                        price[k+1]=price[k];
                        limit[k+1]=limit[k];
                    }
                    price[j]=A;
                    limit[j]=B;
                    break;
                }else if(j==i){
                    price[j]=A;
                    limit[j]=B;
                }
            }
        }
    }
    
    i=0;
    while(M>0){
        if(M>=limit[i]){
            costs+=price[i]*limit[i];
            M-=limit[i];
        }else{
            costs+=price[i]*M;
            M=0;
        }
        i+=1;
    }
    printf("%.lf\n",costs);
    return 0;
}