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>
#include<stdlib.h>

int cmp(const void *, const void *);

int
cmp(const void *a,const void *b){
	if(*(long long *)a>*(long long *)b) return 1;
	if(*(long long *)a=*(long long *)b) return 0;
	if(*(long long *)a<*(long long *)b) return -1;

}

int
main()
{
	int n,m;
	long long a[n];
 	int b[n];
	int i,j;
	
  	scanf("%d %d",&n,&m);
  
	for(i=0;i<n;i++){
   		scanf("%lld %d",&a[i],&b[i]);
      	
  	}

	int sentakushi=0;

	for(i=0;i<n;i++){
		sentakushi+=b[i];
	} 

	
	long long c[sentakushi];
  	int x=0;
  	int k;
	
	for(i=0;i<n;i++){
      	if(i>0){
	      	for(k=0;k<i;k++){
    	      	x += b[k];
        	}
        }
		for(j=0;j<b[i];j++){
			if(i==0){
              c[j]=a[i];
            }else{
              c[j+x]=a[i]; 
            }
		}
	}

	
	qsort((void *)c,sizeof(c)/sizeof(long long),sizeof(long long),cmp);
  	

	
	long long answer=0;
  
	
	for(i=0;i<m;i++){
		answer += c[i];
	}

	printf("%lld",answer);

	return 0;
}