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)

//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
typedef int64_t int64;
typedef uint32_t uint;
typedef uint64_t uint64;
using namespace std;

//---

template <typename T>
inline void print(const T& rhs){ std::cout<<" = "<<rhs<<std::endl; }
template <typename T>
inline void print(const std::vector<T>& rhs){
    std::cout<<" = [ ";
    for(uint i=0; i<rhs.size(); ++i){ std::cout<<rhs[i]<<' '; }
    std::cout<<"]"<<std::endl;
}
template <typename T>
inline void print(const std::vector<std::vector<T>>& rhs){
    std::cout<<" = "<<std::endl;
    std::cout<<"[[ ";
    for(uint p=0; p<rhs.size(); ++p){
        if(p!=0){ std::cout<<" [ "; }
        for(uint q=0; q<rhs[p].size(); ++q){
            std::cout<<rhs[p][q]<<' ';
        }
        if(p!=rhs.size()-1){ std::cout<<"]"<<std::endl; }
    }
    std::cout<<"]]"<<std::endl;
}
template <typename TL, typename TR>
inline void print(const std::vector<std::pair<TR,TL>>& rhs){
    std::cout<<" = [";
    uint i=0;
    for(; i<rhs.size()-1; ++i){ std::cout<<"[f: "<<rhs[i].first<<", s: "<<rhs[i].second<<"], "; }
    std::cout<<"[f: "<<rhs[i].first<<", s: "<<rhs[i].second<<"]]" << endl;
}
#define printn(var) {printf("%s", #var);print(var);}
#define printn_all(var) {printf("%s(%d): ", __func__, __LINE__);printf("%s", #var);print(var);}

//---

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    
    int N, M; cin >> N >> M;
    vector<pair<int,int>> vAB(N);
    for(int i=0; i<N; ++i){
        cin >> vAB[i].first >> vAB[i].second;
    }
    sort(vAB.begin(), vAB.end());

    int64 sum=0;
    int num=0;
    for(int i=0; i<N; ++i){
        if( num+vAB[i].second <= M ){
            num += vAB[i].second;
            sum += (int64)vAB[i].first * (int64)vAB[i].second;
        }else{
            int add_num = M - num;
            if(add_num<0){ break; }
            num += add_num;
            sum += vAB[i].first * add_num;
        }
    }
    cout << sum << endl;
    
    return 0;
}