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.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.IOException;
import java.util.StringTokenizer;
import java.util.List;
import java.util.ArrayList;

public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastScanner fs = new FastScanner(inputStream);
        PrintWriter pw = new PrintWriter(outputStream);
        int N = fs.nextInt(), M = fs.nextInt();
        long minPrice = 0;
        int[] A = new int[N], B = new int[N];
        for(int i = 0; i < N; i++) {
            A[i] = fs.nextInt();
            B[i] = fs.nextInt();
        }
        for(; M > 0 ;) {
            int min = 1000000000;
            int index = 0;
            for (int j = 0; j < N; j++) {
                if(B[j] != 0) min = Math.min(min, A[j]);
            }
            for(int j = 0; j < N; j++) {
                if(A[j] == min && B[j] != 0) {
                    index = j;
                    break;
                }
            }
            if(M <= B[index]) {
                minPrice += M * A[index];
                break;
            } else {
                minPrice += A[index] * B[index];
                M -= B[index];
                B[index] = 0;
            }
        }
        pw.println(minPrice);
        pw.close();
    }
}

class FastScanner {
    private BufferedReader reader = null;
    private StringTokenizer tokenizer = null;

    public FastScanner(InputStream in) {
        reader = new BufferedReader(new InputStreamReader(in));
        tokenizer = null;
    }

    public String next() {
        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                tokenizer = new StringTokenizer(reader.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken();
    }
    public String nextLine() {
        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                return reader.readLine();
            } catch (IOException e) {
                    throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken("\n");
    }

    public long nextLong() {
        return Long.parseLong(next());
    }

    public int nextInt() {
        return Integer.parseInt(next());
    }

    public double nextDouble() {
        return Double.parseDouble(next());
    }

    public int[] nextIntArray(int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++) a[i] = nextInt();
        return a;
    }

    public long[] nextLongArray(int n) {
        long[] a = new long[n];
        for (int i = 0; i < n; i++) a[i] = nextLong();
        return a;
    }

}