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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long LL;
struct X{
	LL A;
	int B;
	X(){
		A=0;
		B=0;
	}
};
bool Comp(X&r,X&l){
	if (r.A<l.A){
		return true;
	}else{
		return false;
	}
}
int main(int argc, char* argv[])
{
	int N,M;
	cin>>N>>M;
	vector<X> AB(N);
	int i;
	for (i=0;i<N;i++){
		cin>>AB[i].A>>AB[i].B;
	}
	sort(AB.begin(),AB.end(),Comp);
	LL ans=0;
	int SumB=0;
	for (i=0;i<N;i++){
		if (SumB==M){
			break;
		}
		if (SumB+AB[i].B<=M){
			SumB+=AB[i].B;
			ans+=AB[i].A*AB[i].B;
		}else{
			int x=M-SumB;
			SumB=M;
			ans+=AB[i].A*x;
		}
	}
	cout<<ans<<endl;	
	return 0;
}