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: Runtime Error)

#include<stdio.h>
#include<algorithm>
using namespace std;
struct node{
	long long s,v;
}Node[100010];
bool cmp(node a,node b){
	return a.v<=b.v;
}
int main(){
	long long n,m,now=0,valu=0;
	scanf("%lld %lld",&n,&m);
	for(int i=0;i<n;i++){
		long long s,v;
		scanf("%lld %lld",&v,&s);
		Node[i].s =s;
		Node[i].v=v;
	}
	sort(Node,Node+n,cmp);
	for(int i=0;i<n;i++){
		if((now+Node[i].s)>=m){
			valu = valu+(m-now)*Node[i].v;
			break;
		}
		now+=Node[i].s;
		valu=valu+Node[i].s*Node[i].v;
	}
	printf("%lld\n",valu);
}