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: Runtime Error)

#include <iostream>
#include <vector>
#include <algorithm>
#define N 100000 //10^5
#define M 100000 //10^5
typedef long long ll;
using namespace std;

int main()
{
    int n, m;
    vector<pair<int, int>> shops;
    ll money = 0;
    int n_drink = 0;

    cin >> n >> m;
    for (int i = 0; i < n; i++)
    {
        int a, b;
        cin >> a >> b;
        shops.push_back(make_pair(a, b));
    }

    sort(shops.begin(), shops.end(), [](pair<int, int> &a, pair<int, int> &b) {
        return a.first < b.first || (a.first == b.first && a.second >= b.second);
    });

    for (int i = 0; i < n; i++)
    {
        for (int j = shops[i].second; j > 0; j--)
        {
            n_drink++;
            money += shops[i].first;
            if (n_drink >= m)
            {
                cout << money << endl;
                return 0;
            }
        }
    }

    return 0;
}