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 <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <cmath>
#include <map>
#include <numeric>
#define FOR(i, a, n) for (int i = (a); i < (n); i++)
#define REP(i, n) for(int i = 0; i < (n); i++)
using namespace std;

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	int N, M;
	long long sum;
	vector<int> drink;

	cin >> N >> M;

	REP(i, N) {
		int A, B;
		cin >> A >> B;
		REP(j, B) drink.push_back(A);
	}

	if (N != 1) sort(drink.begin(), drink.end());

	sum = 0;
	REP(i, M) sum += drink[i];

	cout << sum << endl;

	return 0;
}