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 DATA{
	long long int price;
	long long int bottle;
} DATA;
void Swap(DATA*a,int i,int j){
	DATA tmp = a[i];
	a[i] = a[j];
	a[j] = tmp;
}
int main(void){
	int N,M;
	DATA data[100000];
	scanf("%d %d",&N,&M);
	for(int i = 0;i<N;i++){
		scanf("%d %d",&data[i].price,&data[i].bottle);
	}
	
	for(int i=0;i<N;i++){
		for(int j=i+1;j<N;j++){
			if(data[i].price > data[j].price){
				printf("check %d %d\n",data[i].price,data[j].price);
				Swap(data,i,j);
			}
		}
	}
	printf("\n");
	for(int i=0;i<N;i++){
		printf("%d %d\n",data[i].price,data[i].bottle);
	}
	long long int count =0;
	long long int price = 0;
	for(int i=0;i<N;i++){
		if(count + data[i].bottle > M){
			price += (M-count)*data[i].price;
			break;
		}else{
			price += data[i].price * data[i].bottle;
			count += data[i].bottle;
		}
	}
	printf("%lld\n",price);
	return 0;
}