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(void)
{
	int n,m,a[100000],b[100000],i,j;
	
	scanf("%d %d",&n,&m);
	//printf("n:%d\n", n);
	for(i=0;i<n;i++){
		scanf("%d %d",&a[i],&b[i]);
	}
	//printf("n:%d\n", n);
	int tmp1,tmp2;
	for(i=0;i<n;i++){
		for(j=i+1;j<n;j++){
			if(a[i] > a[j]){
				tmp1 = a[i];
				tmp2 = b[i];
				a[i] = a[j];
				b[i] = b[j];
				a[j] = tmp1;
				b[j] = tmp2;
			}
		}
	}
	//printf("n:%d\n", n);
	//printf("\n");
	for(i=0;i<n;i++){
		//printf("a[%d], b[%d] = %d, %d\n", i, i, a[i], b[i]);
	}
	//printf("n:%d\n", n);
	//printf("\n");
	long long int sum_yen=0,sum_hon=0;
	for(i=0;i<n;i++){
		//printf("i:%d\n",i);
		//printf("在庫あり: a[%d], b[%d] = %d, %d\n", i, i, a[i], b[i]);
		
		sum_hon++;
		b[i]--;
		sum_yen += a[i];
		//printf("sum_hon:%d\n",sum_hon);
		//printf("sum_yen:%d\n",sum_yen);
		if(b[i] == 0){
			//printf("在庫なし: a[%d], b[%d] = %d, %d\n", i, i, a[i], b[i]);
			i++;
		}
		//printf("n:%d\n", n);
		if(sum_hon == m){
			break;
		}
		i--;
	}
	
	
	printf("%lld\n",sum_yen);
	
	
	
	return 0;
}