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: Runtime Error)

#include<stdio.h>
int main(){
    int M,N;
    int A[1024];
    int B[1024];
    float now_m=0;//money
    int now_n=0;//num
    int i;
    float low_p=0;
    scanf("%d %d",&N,&M);
    for(int i = 0; i<N; i++){
        scanf("%d %d",&A[i],&B[i]);
    }
    low_p =10000000000000;

    for(int j=0;j<M;j++){
        for(int i = 0; i<N; i++){
            if(A[i]<low_p && B[i]>0) low_p=A[i];
        }
        for(int i = 0; i<N; i++){
            if(A[i]==low_p && B[i]>0){
 //               printf("a");
                now_m=now_m+A[i];
                B[i]--;
            }
        }
        low_p=10000000000000;;
    }
    printf("%lf\n",now_m);
    return 0;
}