Problem: 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.

Examples

Input

2 5
4 9
2 4


Output

12


Input

4 30
6 18
2 5
3 10
7 9


Output

130


Input

1 100000
1000000000 100000


Output

100000000000000

Solution (in JAVA):

Solution:
import java.util.Map;
import java.util.Map.Entry;
import java.util.Scanner;
import java.util.TreeMap;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = Integer.parseInt(sc.next());
		int M = Integer.parseInt(sc.next());

		Map<Long, Long> m = new TreeMap<>();
		for (int i = 0; i < N; i++) {
			long cost = Long.parseLong(sc.next());
			Long remain = Long.parseLong(sc.next());
			if(m.containsKey(cost)) {
				// 同値はマージしていく
				m.put(cost, (long)m.get(cost) + remain);
			}else {
				m.put(cost, remain);
			}
		}
		
		long ans = solve(N,M,m);
		System.out.println(ans);
	}

	public static long solve(int N, int M, Map<Long, Long> m) {

		long ans = 0;
		int remain = M;
		for (Entry<Long, Long> entry : m.entrySet()) {

			long tmp = remain - entry.getValue();
			if (tmp >= 0) {

				// 1店舗分全部買い取れる
				ans += (long) entry.getKey() * entry.getValue();
			} else {

				// 全部は買い取れない。
				ans += (long) entry.getKey() * remain;
			}
			remain -= entry.getValue();
			if (remain <= 0) {
				// もう買い取る必要がないので抜ける
				break;
			}
		}
		return ans;
	}
}