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: Time Limit Exceeded)

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

#define rep(i,n) for(int i=0;i<(int)(n);i++)

/*Solve*/

int main(){
    ll n,m;
    cin >> n >> m;
    vector<pair<ll,ll>> p;
    rep(i,n){
        ll a,b;
        cin >> a >> b;
        p.push_back({a,b});
    }
    sort(p.begin(),p.end());
    
    ll ans=0;
    while(m>0){
        if(m>=p[0].second){
            ans+=p[0].first*p[0].second;
            m-=p[0].second;
            p.erase(p.begin()+0);
        }else{
            ans+=p[0].first*m;
            m=0;
        }
    }
    cout << ans;
}