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 <bits/stdc++.h>
using namespace std;
const int64_t INF=9999999999999999;

int main() {
   int N,M;
   cin>>N>>M;
   vector<int>A(N);
   vector<int>B(N);

   int64_t ans=0;
   for(int i=0;i<N;i++){
      cin>>A[i]>>B[i];
   }
   vector<pair<int,int>>AB(N);
   for(int i=0;i<N;i++){
      AB[i]=make_pair(A[i],B[i]);
   }
   sort(AB.begin(),AB.end());
   for(int i=0;i<N;i++){
    //  if(c[A[i]]>=M){ans+=A[i]*M;break;}
     // else{ans+=A[i]*c[A[i]];M-=c[A[i]];}
     int64_t buy=min(M,AB[i].second);
     ans+=buy*AB[i].first;
     M-=buy;
   }
   cout<<ans<<endl;
return 0;}