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<cstdio>
#include<algorithm>
using namespace std;
typedef long long LL;
LL X,W;char ch;
inline LL read()
{
	X=0,W=1;ch=getchar();
	while(ch>'9'||ch<'0'){if(ch=='-')W=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){X=(X<<1)+(X<<3)+ch-48;ch=getchar();}
	return X*W;
}
struct store{
	LL num;
	LL pri;
}a[100001];
bool cmp(store c,store b)
{
	return c.pri<b.pri;
}
LL ans;
int main()
{
	int n,m;
	n=read(),m=read();
	for(int i=1;i<=n;i++)a[i].pri=read(),a[i].num=read();
	sort(a+1,a+n+1,cmp);
	int i=1;
	while(true){
		if(a[i].num<m)ans+=a[i].num*a[i].pri;
		else {ans+=m*a[i].pri;break;}
		m-=a[i].num;
		i++;
	}
	printf("%lld",ans);
}