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)

//def
#include <iostream>

#include <cmath>
#include <algorithm>
#include <iomanip>

//array
#include <vector>
#include <set>
#include <map>

using namespace std;
#define rep(i,q) for(int i = 0; (q) > i; i++)
typedef pair<int,int> Paint;
bool Pairfirst(const Paint& pf,Paint&ps){//sortCompare
  return pf.first < ps.first;
}
int main(){
  long long n,m;cin>>n>>m;
  vector<Paint> ss;
  rep(i,n){
    long long tmp,tmpp;cin>>tmp>>tmpp;
    ss.push_back(make_pair(tmp,tmpp));
  }
  sort(ss.begin(),ss.end(),Pairfirst);
  long long  mon = 0;
  long long hon = 0;
  rep(i,n){
    if(m > hon+ss[i].second){
      hon += ss[i].second;
      mon += ss[i].first*ss[i].second;
    }else{
      cout << (long long)mon + (m-hon) * ss[i].first << endl;
      break;
    }
  }
}