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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Map;
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 = sc.nextInt();
		int m = sc.nextInt();
		int[] b = new int[n];
		int[] a = new int[n];
		int[] A = new int[n];
		int[] B = new int[n];
		for(int i = 0; i < n; i++) {
			a[i] = sc.nextInt();
			b[i] = sc.nextInt();
			A[i] = a[i];
			B[i] = b[i];
		}
		Map<Integer, Integer> map = new TreeMap<Integer, Integer>();
		for (int i = 0; i < a.length; i++) {
		    map.put(a[i], b[i]);
		}
		int k = 0;
		for (int nKey: map.keySet()) {
			A[k] = nKey;
			k++;
		}
		k = 0;
		for (int nValue: map.values()) {
			B[k] = nValue;
			k++;
		}
		int cnt = 0;
		long ans = 0;
		int cnt2 = 0;
		int cnt3 = B[0];
		
		while(true) {
			
				if(cnt3 != 0) {
					cnt ++;
					ans += A[cnt2];
					cnt3 --;
				}else {
					cnt2 ++;
					cnt3 = B[cnt2];
				}
				if(cnt == m) {
					break;
				}
			}
			
			
			
		
		System.out.println(ans);
		sc.close();
	}
}