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)

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <cmath>

using namespace std;

struct store
{
	int point, count;
};

bool pred(const store& l, const store& r)
{
	return l.point < r.point;
}

int main()
{
	int N, M;
	cin >> N >> M;

	vector<store> stores(N);

	for (int i = 0; i < N; i++)
	{
		store A;
		cin >> A.point >> A.count;

		stores[i] = A;
	}

	sort(stores.begin(), stores.end(), pred);

	long long ans = 0;
	int index = 0;
	for (int i = 0; i < M; i++)
	{
		if (stores[index].count > 0)
		{
			ans += stores[index].point;
			stores[index].count--;
		}
		else
		{
			index++;
			i--;
		}
	}

	cout << ans << endl;

	return 0;
}