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: Wrong Answer)

#include <stdio.h>

typedef struct{
	int A;
	int B;
}shop_t;

int main(){
	int N,M;
	int i,j;
	int get;
	long long int ans;
	long long int a;
	shop_t m;
	
	scanf("%d %d",&N,&M);
	
	shop_t list[N];
	
	for(i=0;i<N;i++){
		scanf("%d %d",&list[i].A,&list[i].B);
	}
	if(N!=1){
		for(i=0;i<N;i++){
			for(j=i+1;j<N;j++){
				if(list[i].A > list[j].A){
					m=list[i];
					list[i]=list[j];
					list[j]=m;
				}
			}
		}
	}
	
	ans=0;
	get=0;
	i=0;
	while(get!=M){
		while(list[i].B==0)	i++;
		
		if(get+list[i].B>M){
			ans+=(M-get)*list[i].A;
			get=M;
		}else{
			a=list[i].A;
			a*=list[i].B;
			ans+=a;
			get+=list[i].B;
			list[i].B=0;
		}
	}
	
	printf("%lld",ans);
	
	return 0;
}