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: Wrong Answer)

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
using namespace std;
struct juice
{int money;
int number;
}a[100000];
bool cmp(juice a,juice b)
{
    return a.money<b.money;
}
int main()
{long long n,m,sum=0,sum2=0;
cin>>n>>m;
for(int i=0;i<n;i++)
cin>>a[i].money>>a[i].number;
sort(a,a+n,cmp);
for(int i=0;;i++)
{if((sum+a[i].number)<m) {sum2=sum2+a[i].number*a[i].money;sum=sum+a[i].number;}
else if((sum+a[i].number)>=m) {sum2=sum2+(m-sum)*a[i].money;break;}
}
cout<<sum2;
return 0;
}