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)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
#define rep(i,n) for(int i = 0; i < (n); i++)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;

int main() {
  int n,m;
  cin>>n>>m;
  vector<pair<ll,ll>> mz(n);
  rep(i,n){
    ll a,b;
    cin>>a>>b;
    mz[i]=make_pair(a,b);
  }
  sort(mz.begin(),mz.end());
  ll kane=0,zai=0;
  rep(i,n){
    if(zai+mz[i].second<=m){
      kane += mz[i].first*mz[i].second;
      zai += mz[i].second;
    }else if(zai >=m) break;
    else{
      kane += (m-zai)*mz[i].first;
      zai += (m-zai);
    }
  }
  cout<<kane<<endl;
}