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,i,j,count=0;
	long long A[100000],B[100000],min,temp,sum=0;
	
	scanf("%d %d¥n",&N,&M);
	for(i=0;i<N;i++){
		scanf("%ld %ld¥n",&A[i],&B[i]);
	}
	
	/* 最も安い店舗から順に */
	while(count<M){
		min=1000000000;
		for(i=0;i<N;i++){
			if(A[i]<=min){
				min=A[i];
				temp=i;
			}
		}
		count=count+B[temp]; //エナジードリンクの購入数
		sum=sum+A[temp]*B[temp]; //必要金額（最安）
		
		if(count>M){
			sum=sum-A[temp]*(count-M); // 出力時に過剰に購入した分を合計金額から引く
		}
				A[temp]=1000000000;
	}
	printf("%ld",sum);		
	return 0;
}