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++ (GCC 5.4.1) (judge verdict: Wrong Answer)

//ABC121C
#include <iostream>
#include <vector> 
#include <algorithm> 
 
using namespace std;
 
struct Struct {
    int num1;
    int num2;
 
    Struct(int n1, int n2) {
        num1 = n1;
        num2 = n2;
    }
 
    bool operator<(const Struct& another) const {
        return num1 < another.num1;
    }
};

int main() {
    vector<Struct> vec;
    int A[100001]; int B[100001];
    int N,M; cin >> N >> M;
    for(int i = 0;i < N;i++){
        cin >> A[i] >> B[i];
        vec.push_back(Struct(A[i],B[i]));
    }

    sort(vec.begin(),vec.end());
    int sum = 0; int count = 0;
    int j= 0;
    for(j = 0;j < N;j++){
        sum += vec[j].num1 * vec[j].num2;
        count += vec[j].num2;
        if(count >= M){
            sum -= vec[j].num1 * vec[j].num2;
            count -= vec[j].num2;
            break;
        }
    }
    cout << sum + vec[j].num1 * (M - count) << endl;

    return 0;

}